Figure shows four paths for a kicked football. ignoring the eff-Turito

General

Easy

Physics-

Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

Physics-General

4, 3, 2, 1
3, 4, 1, 2
1, 2, 3, 4
2, 3, 4, 1

Answer:The correct answer is: 4, 3, 2, 1 $R equals fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction equals fraction numerator 2 u subscript x end subscript v subscript y end subscript over denominator g end fraction$
$therefore$ Range $proportional to$ horizontal initial velocity $left parenthesis u subscript x end subscript right parenthesis$
In path 4 range is maximum so football possess maximum horizontal velocity in the path

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Related Questions to study

General

physics-

A body of mass $m$ is moving with a uniform speed $v$ along a circle of radius $r$ , what is the average acceleration in going from $A$ to $B$ ?

Here,

T equals fraction numerator 2 pi r over denominator 4 v end fraction equals fraction numerator pi r over denominator 2 v end fraction

Change in velocity is going from

A

B

v square root of 2

Average acceleration

equals fraction numerator v square root of 2 over denominator pi r divided by 2 v end fraction equals fraction numerator 2 square root of 2 v to the power of 2 end exponent over denominator pi r end fraction

A body of mass $m$ is moving with a uniform speed $v$ along a circle of radius $r$ , what is the average acceleration in going from $A$ to $B$ ?

physics-General

Here,

T equals fraction numerator 2 pi r over denominator 4 v end fraction equals fraction numerator pi r over denominator 2 v end fraction

Change in velocity is going from

A

B

v square root of 2

Average acceleration

equals fraction numerator v square root of 2 over denominator pi r divided by 2 v end fraction equals fraction numerator 2 square root of 2 v to the power of 2 end exponent over denominator pi r end fraction

General

physics-

A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

The forces acting on the ball will be (i) in the direction opposite to its motion

i e,

frictional force and(ii) weight

m g

A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

physics-General

The forces acting on the ball will be (i) in the direction opposite to its motion

i e,

frictional force and(ii) weight

m g

General

physics-

A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $theta$ at which the speed of the bob is half of that at $A$ , satisfies

V to the power of 2 end exponent equals U to the power of 2 end exponent minus 2 g left parenthesis L minus L cos invisible function application theta right parenthesis

fraction numerator 5 g L over denominator 4 end fraction equals 5 g L minus 2 g L left parenthesis 1 minus cos invisible function application theta right parenthesis

5 equals 20 minus 8 plus 8 cos invisible function application theta

cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction

fraction numerator 3 pi over denominator 4 end fraction less than theta less than pi

A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $theta$ at which the speed of the bob is half of that at $A$ , satisfies

physics-General

V to the power of 2 end exponent equals U to the power of 2 end exponent minus 2 g left parenthesis L minus L cos invisible function application theta right parenthesis

fraction numerator 5 g L over denominator 4 end fraction equals 5 g L minus 2 g L left parenthesis 1 minus cos invisible function application theta right parenthesis

5 equals 20 minus 8 plus 8 cos invisible function application theta

cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction

fraction numerator 3 pi over denominator 4 end fraction less than theta less than pi

General

physics-

A particle of mass $m$ attracted with a string of length $l$ is just revolving on the vertical circle without slacking of the string. If $v subscript A end subscript comma v subscript B end subscript$ and $v subscript D end subscript$ are speed at position $A comma B$ and $D$ then

A comma blank v subscript A end subscript equals square root of g l end root

B comma blank v subscript B end subscript equals square root of 5 g l end root

and at

D comma blank v subscript D end subscript equals square root of 3 g l end root

Thus,

v subscript B end subscript greater than v subscript D end subscript greater than v subscript A end subscript

Also,

T equals 3 blank m g left parenthesis 1 plus cos invisible function application theta right parenthesis

So,

D comma theta equals 90 degree

therefore blank T equals 3 blank m g open parentheses 1 plus theta close parentheses equals 3 blank m g

A particle of mass $m$ attracted with a string of length $l$ is just revolving on the vertical circle without slacking of the string. If $v subscript A end subscript comma v subscript B end subscript$ and $v subscript D end subscript$ are speed at position $A comma B$ and $D$ then

physics-General

A comma blank v subscript A end subscript equals square root of g l end root

B comma blank v subscript B end subscript equals square root of 5 g l end root

and at

D comma blank v subscript D end subscript equals square root of 3 g l end root

Thus,

v subscript B end subscript greater than v subscript D end subscript greater than v subscript A end subscript

Also,

T equals 3 blank m g left parenthesis 1 plus cos invisible function application theta right parenthesis

So,

D comma theta equals 90 degree

therefore blank T equals 3 blank m g open parentheses 1 plus theta close parentheses equals 3 blank m g

General

physics-

A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $theta$ at which the speed of the bob is half of that at A, satisfies

Velocity of the bob at the point A

v equals square root of 5 g L end root

(i)

open parentheses fraction numerator v over denominator 2 end fraction close parentheses to the power of 2 end exponent equals v to the power of 2 end exponent minus 2 g h open parentheses i i close parentheses

h equals L left parenthesis 1 minus cos invisible function application theta right parenthesis left parenthesis i i i right parenthesis

S o l v i n g blank E q s. open parentheses i close parentheses comma blank open parentheses i i close parentheses a n d blank open parentheses i i i close parentheses comma blank w e blank g e t

cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction

o r blank theta equals c o s to the power of negative 1 end exponent open parentheses negative fraction numerator 7 over denominator 8 end fraction close parentheses equals 151 degree

A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $theta$ at which the speed of the bob is half of that at A, satisfies

physics-General

Velocity of the bob at the point A

v equals square root of 5 g L end root

(i)

open parentheses fraction numerator v over denominator 2 end fraction close parentheses to the power of 2 end exponent equals v to the power of 2 end exponent minus 2 g h open parentheses i i close parentheses

h equals L left parenthesis 1 minus cos invisible function application theta right parenthesis left parenthesis i i i right parenthesis

S o l v i n g blank E q s. open parentheses i close parentheses comma blank open parentheses i i close parentheses a n d blank open parentheses i i i close parentheses comma blank w e blank g e t

cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction

o r blank theta equals c o s to the power of negative 1 end exponent open parentheses negative fraction numerator 7 over denominator 8 end fraction close parentheses equals 151 degree

General

physics-

A projectile $A$ is thrown at an angle of $30 °$ to the horizontal from point $P$ . At the same time, another projectile $B$ is thrown with velocity $v^{2}$ upwards from the point $Q$ vertically below the highest point. For $B$ to collide with $A, \frac{v_{2}}{v_{1}}$ should be

Equating velocities along the vertical,

v_{2} = v_{1} \sin 30 ° o r \frac{v_{2}}{v_{1}} = \frac{1}{2}

A projectile $A$ is thrown at an angle of $30 °$ to the horizontal from point $P$ . At the same time, another projectile $B$ is thrown with velocity $v^{2}$ upwards from the point $Q$ vertically below the highest point. For $B$ to collide with $A, \frac{v_{2}}{v_{1}}$ should be

physics-General

Equating velocities along the vertical,

v_{2} = v_{1} \sin 30 ° o r \frac{v_{2}}{v_{1}} = \frac{1}{2}

General

physics-

In a two dimensional motion of a particle, the particle moves from point $A$ , position vector $stack r with rightwards arrow on top subscript 1 end subscript$ . If the magnitudes of these vectors are respectively, $r subscript 1 end subscript$ =3 and $r subscript 2 end subscript equals 4$ and the angles they make with the $x$ -axis are $theta subscript 1 end subscript equals 75 degree$ and 15 $degree$ , respectively, then find the magnitude of the displacement vector

Displacement

equals A B

angle between

stack r subscript 1 end subscript with rightwards arrow on top

and

stack r subscript 2 end subscript with rightwards arrow on top

theta equals 75 degree minus 15 degree equals 60 degree

From figure

A B to the power of 2 end exponent equals r subscript 1 end subscript superscript 2 end superscript plus r subscript 2 end subscript superscript 2 end superscript minus 2 r subscript 1 end subscript r subscript 2 end subscript c o s theta

equals 3 to the power of 2 end exponent plus 4 to the power of 2 end exponent minus 2 cross times 3 cross times 4

cos

60 degree

equals 13

A B equals square root of 13

In a two dimensional motion of a particle, the particle moves from point $A$ , position vector $stack r with rightwards arrow on top subscript 1 end subscript$ . If the magnitudes of these vectors are respectively, $r subscript 1 end subscript$ =3 and $r subscript 2 end subscript equals 4$ and the angles they make with the $x$ -axis are $theta subscript 1 end subscript equals 75 degree$ and 15 $degree$ , respectively, then find the magnitude of the displacement vector

physics-General

Displacement

equals A B

angle between

stack r subscript 1 end subscript with rightwards arrow on top

and

stack r subscript 2 end subscript with rightwards arrow on top

theta equals 75 degree minus 15 degree equals 60 degree

From figure

A B to the power of 2 end exponent equals r subscript 1 end subscript superscript 2 end superscript plus r subscript 2 end subscript superscript 2 end superscript minus 2 r subscript 1 end subscript r subscript 2 end subscript c o s theta

equals 3 to the power of 2 end exponent plus 4 to the power of 2 end exponent minus 2 cross times 3 cross times 4

cos

60 degree

equals 13

A B equals square root of 13

General

physics-

A particle is moving on a circular path of radius $r$ with uniform velocity $v$ . The change in velocity when the particle moves from $P blank$ to $blank Q blank$ is $blank left parenthesis angle P O Q equals 40 degree right parenthesis$

Change in velocity

equals 2 v sin invisible function application open parentheses theta divided by 2 close parentheses equals 2 v sin invisible function application 20 degree

A particle is moving on a circular path of radius $r$ with uniform velocity $v$ . The change in velocity when the particle moves from $P blank$ to $blank Q blank$ is $blank left parenthesis angle P O Q equals 40 degree right parenthesis$

physics-General

Change in velocity

equals 2 v sin invisible function application open parentheses theta divided by 2 close parentheses equals 2 v sin invisible function application 20 degree

General

physics-

A fighter plane enters inside the enemy territory, at time $t equals 0$ with velocity $v subscript 0 end subscript equals 250 blank m s to the power of negative 1 end exponent$ and moves horizontally with constant acceleration $a equals 20 m s to the power of negative 2 end exponent$ (see figure). An enemy tank at the border, spot the plane and fire shots at an angle $theta equals 60 degree$ with the horizontal and with velocity $u equals 600 blank m s to the power of negative 1 end exponent$ . At what altitude $H$ of the plane it can be hit by the shot?

If it is being hit then

d equals v subscript 0 end subscript t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent equals left parenthesis u cos invisible function application theta right parenthesis t

t equals fraction numerator u cos invisible function application theta minus v subscript 0 end subscript over denominator a divided by 2 end fraction

therefore blank t equals fraction numerator 600 cross times fraction numerator 1 over denominator 2 end fraction minus 250 over denominator 10 end fraction equals 5 blank s

H equals open parentheses u sin invisible function application theta close parentheses t minus fraction numerator 1 over denominator 2 end fraction cross times g t to the power of 2 end exponent

equals 600 cross times fraction numerator square root of 3 over denominator 2 end fraction cross times 5 minus fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 25

H equals 2473 blank m

A fighter plane enters inside the enemy territory, at time $t equals 0$ with velocity $v subscript 0 end subscript equals 250 blank m s to the power of negative 1 end exponent$ and moves horizontally with constant acceleration $a equals 20 m s to the power of negative 2 end exponent$ (see figure). An enemy tank at the border, spot the plane and fire shots at an angle $theta equals 60 degree$ with the horizontal and with velocity $u equals 600 blank m s to the power of negative 1 end exponent$ . At what altitude $H$ of the plane it can be hit by the shot?

physics-General

If it is being hit then

d equals v subscript 0 end subscript t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent equals left parenthesis u cos invisible function application theta right parenthesis t

t equals fraction numerator u cos invisible function application theta minus v subscript 0 end subscript over denominator a divided by 2 end fraction

therefore blank t equals fraction numerator 600 cross times fraction numerator 1 over denominator 2 end fraction minus 250 over denominator 10 end fraction equals 5 blank s

H equals open parentheses u sin invisible function application theta close parentheses t minus fraction numerator 1 over denominator 2 end fraction cross times g t to the power of 2 end exponent

equals 600 cross times fraction numerator square root of 3 over denominator 2 end fraction cross times 5 minus fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 25

H equals 2473 blank m

General

physics-

From an inclined plane two particles are projected with same speed at same angle $theta$ , one up and other down the plane as shown in figure, which of the following statements is/are correct?

H e r e comma blank alpha equals 2 theta comma blank beta equals theta

T i m e blank o f blank f l i g h t blank o f blank A blank i s comma

T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application open parentheses alpha minus beta close parentheses over denominator g cos invisible function application beta end fraction

equals fraction numerator 2 u sin invisible function application open parentheses 2 theta minus theta close parentheses over denominator g cos invisible function application theta end fraction

equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta

T i m e blank o f blank f l i g h t blank o f blank B blank i s comma blank T subscript 2 end subscript equals fraction numerator 2 u sin invisible function application theta over denominator g cos invisible function application theta end fraction

equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta

So,

T subscript 1 end subscript equals T subscript 2 end subscript

. The acceleration of both the particles is

g

downwards. Therefore, relative acceleration between the two I zero or relative motion between the two is uniform. The relative velocity of

A

w.r.t.

B

is towards

A B

, therefore collision will take place between the two in mid air.

From an inclined plane two particles are projected with same speed at same angle $theta$ , one up and other down the plane as shown in figure, which of the following statements is/are correct?

physics-General

H e r e comma blank alpha equals 2 theta comma blank beta equals theta

T i m e blank o f blank f l i g h t blank o f blank A blank i s comma

T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application open parentheses alpha minus beta close parentheses over denominator g cos invisible function application beta end fraction

equals fraction numerator 2 u sin invisible function application open parentheses 2 theta minus theta close parentheses over denominator g cos invisible function application theta end fraction

equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta

T i m e blank o f blank f l i g h t blank o f blank B blank i s comma blank T subscript 2 end subscript equals fraction numerator 2 u sin invisible function application theta over denominator g cos invisible function application theta end fraction

equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta

So,

T subscript 1 end subscript equals T subscript 2 end subscript

. The acceleration of both the particles is

g

downwards. Therefore, relative acceleration between the two I zero or relative motion between the two is uniform. The relative velocity of

A

w.r.t.

B

is towards

A B

, therefore collision will take place between the two in mid air.

General

physics-

A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2 divided by pi$ revolutions per $s e c o n d$ around the vertical axis through the fixed end as shown in figure , then tension in the string is

T sin invisible function application theta equals M omega to the power of 2 end exponent R

(i)

T sin invisible function application theta equals M omega to the power of 2 end exponent L sin invisible function application theta

(ii)
From (i) and (ii)

T equals M omega to the power of 2 end exponent L

equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L

equals M blank 4 pi to the power of 2 end exponent open parentheses fraction numerator 2 over denominator pi end fraction close parentheses to the power of 2 end exponent L equals 16 blank M L

A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2 divided by pi$ revolutions per $s e c o n d$ around the vertical axis through the fixed end as shown in figure , then tension in the string is

physics-General

T sin invisible function application theta equals M omega to the power of 2 end exponent R

(i)

T sin invisible function application theta equals M omega to the power of 2 end exponent L sin invisible function application theta

(ii)
From (i) and (ii)

T equals M omega to the power of 2 end exponent L

equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L

equals M blank 4 pi to the power of 2 end exponent open parentheses fraction numerator 2 over denominator pi end fraction close parentheses to the power of 2 end exponent L equals 16 blank M L

General

physics-

A piece of wire is bent in the shape of a parabola $y equals k x to the power of 2 end exponent blank left parenthesis y$ -axis vertical) with a bead of mass $m$ on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$ -axis with a constant acceleration $a$ . The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the $y$ -axis is

m a cos invisible function application theta equals m g cos invisible function application left parenthesis 90 minus theta right parenthesis

rightwards double arrow fraction numerator a over denominator g end fraction equals tan invisible function application theta rightwards double arrow fraction numerator a over denominator g end fraction equals fraction numerator d y over denominator d x end fraction

rightwards double arrow fraction numerator d over denominator d x end fraction open parentheses k x close parentheses to the power of 2 end exponent equals fraction numerator a over denominator g end fraction rightwards double arrow x equals fraction numerator a over denominator 2 g k end fraction

A piece of wire is bent in the shape of a parabola $y equals k x to the power of 2 end exponent blank left parenthesis y$ -axis vertical) with a bead of mass $m$ on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$ -axis with a constant acceleration $a$ . The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the $y$ -axis is

physics-General

m a cos invisible function application theta equals m g cos invisible function application left parenthesis 90 minus theta right parenthesis

rightwards double arrow fraction numerator a over denominator g end fraction equals tan invisible function application theta rightwards double arrow fraction numerator a over denominator g end fraction equals fraction numerator d y over denominator d x end fraction

rightwards double arrow fraction numerator d over denominator d x end fraction open parentheses k x close parentheses to the power of 2 end exponent equals fraction numerator a over denominator g end fraction rightwards double arrow x equals fraction numerator a over denominator 2 g k end fraction

General

physics-

A frictionless track $A B C D E$ ends in a circular loop of radius $R comma$ figure. A body slides down the track from point $A$ which is at a height $h equals 5 blank$ cm. Maximum value of $R$ for the body to successfully complete the loop is

For successfully completing the loop,

h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m

A frictionless track $A B C D E$ ends in a circular loop of radius $R comma$ figure. A body slides down the track from point $A$ which is at a height $h equals 5 blank$ cm. Maximum value of $R$ for the body to successfully complete the loop is

physics-General

For successfully completing the loop,

h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m

General

physics-

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through $R$ , where $P R$ equals

Equating the moments about

R

6 cross times P R equals 4 cross times R Q

P R equals fraction numerator 4 over denominator 6 end fraction R Q equals fraction numerator 2 over denominator 3 end fraction R Q

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through $R$ , where $P R$ equals

physics-General

Equating the moments about

R

6 cross times P R equals 4 cross times R Q

P R equals fraction numerator 4 over denominator 6 end fraction R Q equals fraction numerator 2 over denominator 3 end fraction R Q

General

physics-

The time taken by the projectile to reach from $A$ to $B$ is $t comma$ then the distance $A B$ is equal to

Horizontal component of velocity at

A

v subscript H end subscript equals u cos invisible function application 60 degree equals fraction numerator u over denominator 2 end fraction blank therefore A C equals u subscript H end subscript cross times t equals fraction numerator u t over denominator 2 end fraction

A B equals A C sec invisible function application 30 degree equals fraction numerator u t over denominator 2 end fraction cross times fraction numerator 2 over denominator square root of 3 end fraction equals fraction numerator u t over denominator 2 end fraction

The time taken by the projectile to reach from $A$ to $B$ is $t comma$ then the distance $A B$ is equal to

physics-General

Horizontal component of velocity at

A

v subscript H end subscript equals u cos invisible function application 60 degree equals fraction numerator u over denominator 2 end fraction blank therefore A C equals u subscript H end subscript cross times t equals fraction numerator u t over denominator 2 end fraction

A B equals A C sec invisible function application 30 degree equals fraction numerator u t over denominator 2 end fraction cross times fraction numerator 2 over denominator square root of 3 end fraction equals fraction numerator u t over denominator 2 end fraction

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Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

4, 3, 2, 1
3, 4, 1, 2
1, 2, 3, 4
2, 3, 4, 1

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Related Questions to study

A body of mass $m$ is moving with a uniform speed $v$ along a circle of radius $r$ , what is the average acceleration in going from $A$ to $B$ ?

A body of mass $m$ is moving with a uniform speed $v$ along a circle of radius $r$ , what is the average acceleration in going from $A$ to $B$ ?

A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

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A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $theta$ at which the speed of the bob is half of that at $A$ , satisfies

A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $theta$ at which the speed of the bob is half of that at $A$ , satisfies

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A particle of mass $m$ attracted with a string of length $l$ is just revolving on the vertical circle without slacking of the string. If $v subscript A end subscript comma v subscript B end subscript$ and $v subscript D end subscript$ are speed at position $A comma B$ and $D$ then

A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $theta$ at which the speed of the bob is half of that at A, satisfies

A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $theta$ at which the speed of the bob is half of that at A, satisfies

A projectile $A$ is thrown at an angle of $30 °$ to the horizontal from point $P$ . At the same time, another projectile $B$ is thrown with velocity $v^{2}$ upwards from the point $Q$ vertically below the highest point. For $B$ to collide with $A, \frac{v_{2}}{v_{1}}$ should be

A projectile $A$ is thrown at an angle of $30 °$ to the horizontal from point $P$ . At the same time, another projectile $B$ is thrown with velocity $v^{2}$ upwards from the point $Q$ vertically below the highest point. For $B$ to collide with $A, \frac{v_{2}}{v_{1}}$ should be

A particle is moving on a circular path of radius $r$ with uniform velocity $v$ . The change in velocity when the particle moves from $P blank$ to $blank Q blank$ is $blank left parenthesis angle P O Q equals 40 degree right parenthesis$

A particle is moving on a circular path of radius $r$ with uniform velocity $v$ . The change in velocity when the particle moves from $P blank$ to $blank Q blank$ is $blank left parenthesis angle P O Q equals 40 degree right parenthesis$

From an inclined plane two particles are projected with same speed at same angle $theta$ , one up and other down the plane as shown in figure, which of the following statements is/are correct?

From an inclined plane two particles are projected with same speed at same angle $theta$ , one up and other down the plane as shown in figure, which of the following statements is/are correct?

A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2 divided by pi$ revolutions per $s e c o n d$ around the vertical axis through the fixed end as shown in figure , then tension in the string is

A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2 divided by pi$ revolutions per $s e c o n d$ around the vertical axis through the fixed end as shown in figure , then tension in the string is

A frictionless track $A B C D E$ ends in a circular loop of radius $R comma$ figure. A body slides down the track from point $A$ which is at a height $h equals 5 blank$ cm. Maximum value of $R$ for the body to successfully complete the loop is

A frictionless track $A B C D E$ ends in a circular loop of radius $R comma$ figure. A body slides down the track from point $A$ which is at a height $h equals 5 blank$ cm. Maximum value of $R$ for the body to successfully complete the loop is

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through $R$ , where $P R$ equals

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through $R$ , where $P R$ equals

The time taken by the projectile to reach from $A$ to $B$ is $t comma$ then the distance $A B$ is equal to

The time taken by the projectile to reach from $A$ to $B$ is $t comma$ then the distance $A B$ is equal to

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Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

4, 3, 2, 1 3, 4, 1, 2 1, 2, 3, 4 2, 3, 4, 1

Answer:The correct answer is: 4, 3, 2, 1 Range horizontal initial velocity In path 4 range is maximum so football possess maximum horizontal velocity in the path

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Related Questions to study

A body of mass is moving with a uniform speed along a circle of radius , what is the average acceleration in going from to ?

A body of mass is moving with a uniform speed along a circle of radius , what is the average acceleration in going from to ?

A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

A bob of mass is suspended by a massless string of length . The horizontal velocity at position is just sufficient to make it reach the point . The angle at which the speed of the bob is half of that at , satisfies

A bob of mass is suspended by a massless string of length . The horizontal velocity at position is just sufficient to make it reach the point . The angle at which the speed of the bob is half of that at , satisfies

A particle of mass attracted with a string of length is just revolving on the vertical circle without slacking of the string. If and are speed at position and then

A particle of mass attracted with a string of length is just revolving on the vertical circle without slacking of the string. If and are speed at position and then

A bob of mass M is suspended by a massless string of length L. The horizontal velocity at position A is just sufficient to make it reach the point B. The angle at which the speed of the bob is half of that at A, satisfies

A bob of mass M is suspended by a massless string of length L. The horizontal velocity at position A is just sufficient to make it reach the point B. The angle at which the speed of the bob is half of that at A, satisfies

A projectile A is thrown at an angle of 30° to the horizontal from point P. At the same time, another projectile B is thrown with velocity v2 upwards from the point Q vertically below the highest point. For B to collide with A, v2v1 should be

A projectile A is thrown at an angle of 30° to the horizontal from point P. At the same time, another projectile B is thrown with velocity v2 upwards from the point Q vertically below the highest point. For B to collide with A, v2v1 should be

In a two dimensional motion of a particle, the particle moves from point , position vector . If the magnitudes of these vectors are respectively, =3 and and the angles they make with the -axis are and 15, respectively, then find the magnitude of the displacement vector

In a two dimensional motion of a particle, the particle moves from point , position vector . If the magnitudes of these vectors are respectively, =3 and and the angles they make with the -axis are and 15, respectively, then find the magnitude of the displacement vector

A particle is moving on a circular path of radius with uniform velocity . The change in velocity when the particle moves from tois

A particle is moving on a circular path of radius with uniform velocity . The change in velocity when the particle moves from tois

From an inclined plane two particles are projected with same speed at same angle , one up and other down the plane as shown in figure, which of the following statements is/are correct?

From an inclined plane two particles are projected with same speed at same angle , one up and other down the plane as shown in figure, which of the following statements is/are correct?

A string of length is fixed at one end and carries a mass at the other end. The string makes revolutions per around the vertical axis through the fixed end as shown in figure , then tension in the string is

A string of length is fixed at one end and carries a mass at the other end. The string makes revolutions per around the vertical axis through the fixed end as shown in figure , then tension in the string is

A frictionless track ends in a circular loop of radius figure. A body slides down the track from point which is at a height cm. Maximum value of for the body to successfully complete the loop is

A frictionless track ends in a circular loop of radius figure. A body slides down the track from point which is at a height cm. Maximum value of for the body to successfully complete the loop is

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through , where equals

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through , where equals

The time taken by the projectile to reach from to is then the distance is equal to

The time taken by the projectile to reach from to is then the distance is equal to

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A body of mass $m$ is moving with a uniform speed $v$ along a circle of radius $r$ , what is the average acceleration in going from $A$ to $B$ ?

A body of mass $m$ is moving with a uniform speed $v$ along a circle of radius $r$ , what is the average acceleration in going from $A$ to $B$ ?

A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $theta$ at which the speed of the bob is half of that at $A$ , satisfies

A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $theta$ at which the speed of the bob is half of that at $A$ , satisfies

A particle of mass $m$ attracted with a string of length $l$ is just revolving on the vertical circle without slacking of the string. If $v subscript A end subscript comma v subscript B end subscript$ and $v subscript D end subscript$ are speed at position $A comma B$ and $D$ then

A particle of mass $m$ attracted with a string of length $l$ is just revolving on the vertical circle without slacking of the string. If $v subscript A end subscript comma v subscript B end subscript$ and $v subscript D end subscript$ are speed at position $A comma B$ and $D$ then

A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $theta$ at which the speed of the bob is half of that at A, satisfies

A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $theta$ at which the speed of the bob is half of that at A, satisfies

A projectile $A$ is thrown at an angle of $30 °$ to the horizontal from point $P$ . At the same time, another projectile $B$ is thrown with velocity $v^{2}$ upwards from the point $Q$ vertically below the highest point. For $B$ to collide with $A, \frac{v_{2}}{v_{1}}$ should be

A projectile $A$ is thrown at an angle of $30 °$ to the horizontal from point $P$ . At the same time, another projectile $B$ is thrown with velocity $v^{2}$ upwards from the point $Q$ vertically below the highest point. For $B$ to collide with $A, \frac{v_{2}}{v_{1}}$ should be

A particle is moving on a circular path of radius $r$ with uniform velocity $v$ . The change in velocity when the particle moves from $P blank$ to $blank Q blank$ is $blank left parenthesis angle P O Q equals 40 degree right parenthesis$

A particle is moving on a circular path of radius $r$ with uniform velocity $v$ . The change in velocity when the particle moves from $P blank$ to $blank Q blank$ is $blank left parenthesis angle P O Q equals 40 degree right parenthesis$

From an inclined plane two particles are projected with same speed at same angle $theta$ , one up and other down the plane as shown in figure, which of the following statements is/are correct?

From an inclined plane two particles are projected with same speed at same angle $theta$ , one up and other down the plane as shown in figure, which of the following statements is/are correct?

A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2 divided by pi$ revolutions per $s e c o n d$ around the vertical axis through the fixed end as shown in figure , then tension in the string is

A string of length $L$ is fixed at one end and carries a mass $M$ at the other end. The string makes $2 divided by pi$ revolutions per $s e c o n d$ around the vertical axis through the fixed end as shown in figure , then tension in the string is

A frictionless track $A B C D E$ ends in a circular loop of radius $R comma$ figure. A body slides down the track from point $A$ which is at a height $h equals 5 blank$ cm. Maximum value of $R$ for the body to successfully complete the loop is

A frictionless track $A B C D E$ ends in a circular loop of radius $R comma$ figure. A body slides down the track from point $A$ which is at a height $h equals 5 blank$ cm. Maximum value of $R$ for the body to successfully complete the loop is

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through $R$ , where $P R$ equals

The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through $R$ , where $P R$ equals

The time taken by the projectile to reach from $A$ to $B$ is $t comma$ then the distance $A B$ is equal to

The time taken by the projectile to reach from $A$ to $B$ is $t comma$ then the distance $A B$ is equal to