General
Easy
Physics

Angle of projection, maximum height and time to reach the maximum height of a particle are theta,H and  respectivley. Find the true relation.

PhysicsGeneral

  1. tm=square root of fraction numerator 4 H over denominator g end fraction end root
  2. tm=square root of fraction numerator H over denominator 4 g end fraction end root
  3. tm=square root of fraction numerator 2 H over denominator g end fraction end root
  4. tm=square root of fraction numerator H over denominator 2 g end fraction end root

    Answer:The correct answer is: tm=square root of fraction numerator 2 H over denominator g end fraction end root

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    Related Questions to study

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    A particle A is projected from the ground with an initial velocity of 10 blank m s to the power of negative 1 end exponentat an angle of 60degree with horizontal. From what height h should an another particle B be projected horizontally with velocity 5 m s to the power of negative 1 end exponent so that both the particles collide in ground at point C if both are projected simultaneously? left parenthesis g equals 10 blank m s to the power of negative 2 end exponent right parenthesis

    Horizontal component of velocity of A is 10 c o s blank 6 0 degree or 5 blank m s to the power of negative 1 end exponent which is equal to the velocity of B in horizontal direction. They will collide at C if time of flight of the particles are equal or t subscript A end subscript equals t subscript B end subscript
    fraction numerator 2 u sin invisible function application theta over denominator g end fraction equals square root of fraction numerator 2 h over denominator g end fraction end root open parentheses therefore h equals fraction numerator 1 over denominator 2 end fraction g t subscript B end subscript superscript 2 end superscript close parentheses
    or h equals fraction numerator 2 u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application theta over denominator g end fraction
    equals fraction numerator 2 open parentheses 10 close parentheses to the power of 2 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses to the power of 2 end exponent over denominator 10 end fraction equals 15 blank m

    A particle A is projected from the ground with an initial velocity of 10 blank m s to the power of negative 1 end exponentat an angle of 60degree with horizontal. From what height h should an another particle B be projected horizontally with velocity 5 m s to the power of negative 1 end exponent so that both the particles collide in ground at point C if both are projected simultaneously? left parenthesis g equals 10 blank m s to the power of negative 2 end exponent right parenthesis

    physics-General
    Horizontal component of velocity of A is 10 c o s blank 6 0 degree or 5 blank m s to the power of negative 1 end exponent which is equal to the velocity of B in horizontal direction. They will collide at C if time of flight of the particles are equal or t subscript A end subscript equals t subscript B end subscript
    fraction numerator 2 u sin invisible function application theta over denominator g end fraction equals square root of fraction numerator 2 h over denominator g end fraction end root open parentheses therefore h equals fraction numerator 1 over denominator 2 end fraction g t subscript B end subscript superscript 2 end superscript close parentheses
    or h equals fraction numerator 2 u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application theta over denominator g end fraction
    equals fraction numerator 2 open parentheses 10 close parentheses to the power of 2 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses to the power of 2 end exponent over denominator 10 end fraction equals 15 blank m
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    A particle is projected with a speed v at 45 degree with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

    Velocity of particle at maximum height h is v to the power of ´ end exponent equals v cos invisible function application theta where v equals initial velocity of particle at which it is projected, theta equals angle of projection
    Angular momentum, L equals m v to the power of ´ end exponent h equals m v cos invisible function application theta h
    equals m v h cos invisible function application 45 degree equals fraction numerator m v h over denominator square root of 2 end fraction

    A particle is projected with a speed v at 45 degree with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

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    Velocity of particle at maximum height h is v to the power of ´ end exponent equals v cos invisible function application theta where v equals initial velocity of particle at which it is projected, theta equals angle of projection
    Angular momentum, L equals m v to the power of ´ end exponent h equals m v cos invisible function application theta h
    equals m v h cos invisible function application 45 degree equals fraction numerator m v h over denominator square root of 2 end fraction
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    A body of mass 1kg is rotating in a vertical circle of radius 1m. What will be the difference in its kinetic energy at the top and bottom of the circle? (Take g equals 10 blank m s to the power of negative 2 end exponent)

    Difference in K E equals fraction numerator 1 over denominator 2 end fraction m open square brackets open parentheses square root of 5 g r end root close parentheses to the power of 2 end exponent minus square root of g r end root close square brackets to the power of 2 end exponent
    equals 2 m g r equals 2 cross times 1 cross times 10 cross times 1 equals 20 blankJ

    A body of mass 1kg is rotating in a vertical circle of radius 1m. What will be the difference in its kinetic energy at the top and bottom of the circle? (Take g equals 10 blank m s to the power of negative 2 end exponent)

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    Difference in K E equals fraction numerator 1 over denominator 2 end fraction m open square brackets open parentheses square root of 5 g r end root close parentheses to the power of 2 end exponent minus square root of g r end root close square brackets to the power of 2 end exponent
    equals 2 m g r equals 2 cross times 1 cross times 10 cross times 1 equals 20 blankJ
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    A particle of mass m moves with constant speed along a circular path of radius r under the action of force F. Its speed is

    F equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction or v equals square root of fraction numerator F r over denominator m end fraction end root

    A particle of mass m moves with constant speed along a circular path of radius r under the action of force F. Its speed is

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    F equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction or v equals square root of fraction numerator F r over denominator m end fraction end root
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    Two stones are projected with the same velocity in magnitude but making different angles with the horizontal. Their ranges are equal. If the angel of projection of one is pi divided by 3 and its maximum height is y subscript 1 end subscript, the maximum height of the other will be

    Given theta subscript 1 end subscript equals pi divided by 3 equals 30 degree
    Horizontal range is same if theta subscript 1 end subscript plus theta subscript 2 end subscript equals 90 degree
    therefore theta subscript 2 end subscript equals 90 degree minus 30 degree equals 60 degree
    y subscript 1 end subscript equals fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 30 degree over denominator 2 g end fraction and y subscript 2 end subscript fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 60 degree over denominator 2 g end fraction
    therefore fraction numerator y subscript 2 end subscript over denominator y subscript 1 end subscript end fraction equals fraction numerator sin to the power of 2 end exponent invisible function application 30 degree over denominator sin to the power of 2 end exponent invisible function application 60 degree end fraction equals open parentheses fraction numerator 1 divided by 4 over denominator square root of 3 divided by 4 end fraction close parentheses to the power of 2 end exponent blank equals fraction numerator 1 over denominator 2 end fraction blank o r blank y subscript 2 end subscript equals fraction numerator y subscript 1 end subscript over denominator 3 end fraction

    Two stones are projected with the same velocity in magnitude but making different angles with the horizontal. Their ranges are equal. If the angel of projection of one is pi divided by 3 and its maximum height is y subscript 1 end subscript, the maximum height of the other will be

    physics-General
    Given theta subscript 1 end subscript equals pi divided by 3 equals 30 degree
    Horizontal range is same if theta subscript 1 end subscript plus theta subscript 2 end subscript equals 90 degree
    therefore theta subscript 2 end subscript equals 90 degree minus 30 degree equals 60 degree
    y subscript 1 end subscript equals fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 30 degree over denominator 2 g end fraction and y subscript 2 end subscript fraction numerator u to the power of 2 end exponent sin to the power of 2 end exponent invisible function application 60 degree over denominator 2 g end fraction
    therefore fraction numerator y subscript 2 end subscript over denominator y subscript 1 end subscript end fraction equals fraction numerator sin to the power of 2 end exponent invisible function application 30 degree over denominator sin to the power of 2 end exponent invisible function application 60 degree end fraction equals open parentheses fraction numerator 1 divided by 4 over denominator square root of 3 divided by 4 end fraction close parentheses to the power of 2 end exponent blank equals fraction numerator 1 over denominator 2 end fraction blank o r blank y subscript 2 end subscript equals fraction numerator y subscript 1 end subscript over denominator 3 end fraction
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    Two wires A C and B C are tied at C of small sphere of mass 5 kg, which revolves at a constant speed v in the horizontal circle of radius 1.6 m. The minimum value of v is

    From force diagram shown in figure
    T subscript 1 end subscript cos invisible function application 30 degree plus T subscript 2 end subscript cos invisible function application 45 degree equals m g (i)
    T subscript 1 end subscript sin invisible function application 30 degree plus T subscript 2 end subscript sin invisible function application 45 degree equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction (ii)
    After solving Eq. (i) and eq. (ii), we get
    T subscript 1 end subscript equals fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator open parentheses fraction numerator square root of 3 minus 1 over denominator 2 end fraction close parentheses end fraction
    But T subscript 1 end subscript greater than 0
    therefore fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator fraction numerator square root of 3 minus 1 over denominator 2 end fraction end fraction greater than 0

    or m g greater than fraction numerator m v to the power of 2 end exponent over denominator r end fraction
    or v less than square root of r g end root
    therefore v subscript m a x end subscript equals square root of r g end root equals square root of 1.6 cross times 9.8 end root equals 3.96 blank m s to the power of negative 1 end exponent

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    From force diagram shown in figure
    T subscript 1 end subscript cos invisible function application 30 degree plus T subscript 2 end subscript cos invisible function application 45 degree equals m g (i)
    T subscript 1 end subscript sin invisible function application 30 degree plus T subscript 2 end subscript sin invisible function application 45 degree equals fraction numerator m v to the power of 2 end exponent over denominator r end fraction (ii)
    After solving Eq. (i) and eq. (ii), we get
    T subscript 1 end subscript equals fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator open parentheses fraction numerator square root of 3 minus 1 over denominator 2 end fraction close parentheses end fraction
    But T subscript 1 end subscript greater than 0
    therefore fraction numerator m g minus fraction numerator m v to the power of 2 end exponent over denominator r end fraction over denominator fraction numerator square root of 3 minus 1 over denominator 2 end fraction end fraction greater than 0

    or m g greater than fraction numerator m v to the power of 2 end exponent over denominator r end fraction
    or v less than square root of r g end root
    therefore v subscript m a x end subscript equals square root of r g end root equals square root of 1.6 cross times 9.8 end root equals 3.96 blank m s to the power of negative 1 end exponent
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    physics-

    The horizontal range of a projectile 4 square root of 3 times the maximum height achieved by it, then the angle of projection is

    fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction equals 4 square root of 3 cross times fraction numerator u to the power of 2 end exponent sin invisible function application theta over denominator 2 g end fraction
    or fraction numerator u to the power of 2 end exponent over denominator g end fraction 2 sin invisible function application theta cos invisible function application theta equals 2 square root of 3 fraction numerator u to the power of 2 end exponent over denominator g end fraction sin to the power of 2 end exponent invisible function application theta

    The horizontal range of a projectile 4 square root of 3 times the maximum height achieved by it, then the angle of projection is

    physics-General
    fraction numerator u to the power of 2 end exponent sin invisible function application 2 theta over denominator g end fraction equals 4 square root of 3 cross times fraction numerator u to the power of 2 end exponent sin invisible function application theta over denominator 2 g end fraction
    or fraction numerator u to the power of 2 end exponent over denominator g end fraction 2 sin invisible function application theta cos invisible function application theta equals 2 square root of 3 fraction numerator u to the power of 2 end exponent over denominator g end fraction sin to the power of 2 end exponent invisible function application theta
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    physics-

    An object is moving in a circle of radius 100 m with a constant speed of 31.4 blank m s to the power of negative 1 end exponent. What is its average speed for one complete revolution?

    The time taken by the particle for one complete revolution.
    t equals fraction numerator 2 pi r over denominator s p e e d end fraction
    equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 31.4 end fraction equals 20 s
    H e n c e comma blank a v e r g e blank s p e e d blank i s blank
    v subscript a v end subscript equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 20 end fraction equals 31.4 blank m s to the power of negative 1 end exponent

    An object is moving in a circle of radius 100 m with a constant speed of 31.4 blank m s to the power of negative 1 end exponent. What is its average speed for one complete revolution?

    physics-General
    The time taken by the particle for one complete revolution.
    t equals fraction numerator 2 pi r over denominator s p e e d end fraction
    equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 31.4 end fraction equals 20 s
    H e n c e comma blank a v e r g e blank s p e e d blank i s blank
    v subscript a v end subscript equals fraction numerator 2 cross times 3.14 cross times 100 over denominator 20 end fraction equals 31.4 blank m s to the power of negative 1 end exponent
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    In the formation of homo diatomic neutral molecule, if ‘N’ atomic orbitals combine, then the total number of bonding molecular orbitals formed is

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    A particle is moving in a circle of radius R with constant speed. It coveres an angle theta in some time interval. Find displacement in this interval of time.

    A particle is moving in a circle of radius R with constant speed. It coveres an angle theta in some time interval. Find displacement in this interval of time.

    physicsGeneral
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    A circular road of radius 1000 blank m has banking angle 45 degree. The maximum safe speed of a car having mass 2000 blank k g will be, if the coefficient of friction between tyre and road is 0.5

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    v to the power of 2 end exponent equals g r open parentheses fraction numerator u plus t a n theta over denominator 1 minus u t a n theta end fraction close parentheses
    rightwards double arrow v to the power of 2 end exponent equals 9.8 cross times 1000 cross times open parentheses fraction numerator 0.5 plus 1 over denominator 1 minus 0.5 cross times 1 end fraction close parentheses rightwards double arrow v equals 172 m divided by s

    A circular road of radius 1000 blank m has banking angle 45 degree. The maximum safe speed of a car having mass 2000 blank k g will be, if the coefficient of friction between tyre and road is 0.5

    physics-General
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    v to the power of 2 end exponent equals g r open parentheses fraction numerator u plus t a n theta over denominator 1 minus u t a n theta end fraction close parentheses
    rightwards double arrow v to the power of 2 end exponent equals 9.8 cross times 1000 cross times open parentheses fraction numerator 0.5 plus 1 over denominator 1 minus 0.5 cross times 1 end fraction close parentheses rightwards double arrow v equals 172 m divided by s