General
Easy
Physics-

A man standing on a hill top projects a stone horizontally with speed v subscript 0 end subscript as shown in figure. Taking the coordinate system as given in the figure. The coordinates of the point where the stone will hit the hill surface

Physics-General

  1. open parentheses fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript t a n theta over denominator g end fraction comma fraction numerator negative 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction close parentheses    
  2. open parentheses fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript over denominator g end fraction comma fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction close parentheses    
  3. open parentheses fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan invisible function application theta over denominator g end fraction comma fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript over denominator g end fraction close parentheses    
  4. open parentheses fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction comma fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan invisible function application theta over denominator g end fraction close parentheses    

    Answer:The correct answer is: open parentheses fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript t a n theta over denominator g end fraction comma fraction numerator negative 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction close parenthesesRange of the projectile on an inclined plane (down the plane) is,
    R equals fraction numerator u to the power of 2 end exponent over denominator g c o s to the power of 2 end exponent beta end fraction left square bracket sin invisible function application open parentheses 2 alpha plus beta close parentheses plus sin invisible function application beta right square bracket
    Here, u equals v subscript 0 end subscript comma alpha equals 0 and beta equals theta
    therefore R equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript sin invisible function application theta over denominator g cos to the power of 2 end exponent invisible function application theta end fraction

    Now x equals R cos invisible function application theta equals fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan invisible function application theta over denominator g end fraction
    and y equals negative R sin invisible function application theta equals blank minus fraction numerator 2 v subscript 0 end subscript superscript 2 end superscript tan to the power of 2 end exponent invisible function application theta over denominator g end fraction

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    Related Questions to study

    General
    physics-

    A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at x equals 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x equals 16 blank m is

    Work done W equals Area under F-xgraph with proper sign W equals Area of triangle A B C + Area of rectangle C D E F + Area of rectangle F G H I + Area of I J K L

    W equals open square brackets fraction numerator 1 over denominator 2 end fraction blank cross times 6 cross times 10 close square brackets plus open square brackets 4 cross times left parenthesis negative 5 right parenthesis close square brackets plus open square brackets 4 cross times 5 close square brackets plus open square brackets 2 cross times open parentheses negative 5 close parentheses close square brackets
    rightwards double arrow W equals 30 minus 20 plus 20 minus 10 equals 20 blank J ….(i)
    According to work energy theorem
    K subscript f end subscript minus K subscript i end subscript equals W rightwards double arrow open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript minus open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript equals W
    open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript equals open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript plus W
    equals 25 blank J plus 20 blank J equals 45 blank J [Using (i)]

    A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at x equals 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x equals 16 blank m is

    physics-General
    Work done W equals Area under F-xgraph with proper sign W equals Area of triangle A B C + Area of rectangle C D E F + Area of rectangle F G H I + Area of I J K L

    W equals open square brackets fraction numerator 1 over denominator 2 end fraction blank cross times 6 cross times 10 close square brackets plus open square brackets 4 cross times left parenthesis negative 5 right parenthesis close square brackets plus open square brackets 4 cross times 5 close square brackets plus open square brackets 2 cross times open parentheses negative 5 close parentheses close square brackets
    rightwards double arrow W equals 30 minus 20 plus 20 minus 10 equals 20 blank J ….(i)
    According to work energy theorem
    K subscript f end subscript minus K subscript i end subscript equals W rightwards double arrow open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript minus open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript equals W
    open parentheses K subscript f end subscript close parentheses subscript x equals 16 m end subscript equals open parentheses K subscript i end subscript close parentheses subscript x equals 0 m end subscript plus W
    equals 25 blank J plus 20 blank J equals 45 blank J [Using (i)]
    General
    physics-

    A bob of mass m accelerates uniformly from rest to v subscript 1 end subscript in time t subscript 1 end subscript. As a function of t, the instantaneous power delivered to the body is

    From v equals u plus a t comma v subscript 1 end subscript equals 0 plus a t subscript 1 end subscript open parentheses because a equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction close parentheses
    F equals m a equals m v subscript 1 end subscript divided by t subscript 1 end subscript
    Velocity acquired in t sec equals a t equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction t
    Power equals F cross times v equals fraction numerator m v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction cross times fraction numerator v subscript 1 end subscript t over denominator t subscript 1 end subscript end fraction equals fraction numerator m v subscript 1 end subscript superscript 2 end superscript t over denominator t subscript 1 end subscript superscript 2 end superscript end fraction

    A bob of mass m accelerates uniformly from rest to v subscript 1 end subscript in time t subscript 1 end subscript. As a function of t, the instantaneous power delivered to the body is

    physics-General
    From v equals u plus a t comma v subscript 1 end subscript equals 0 plus a t subscript 1 end subscript open parentheses because a equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction close parentheses
    F equals m a equals m v subscript 1 end subscript divided by t subscript 1 end subscript
    Velocity acquired in t sec equals a t equals fraction numerator v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction t
    Power equals F cross times v equals fraction numerator m v subscript 1 end subscript over denominator t subscript 1 end subscript end fraction cross times fraction numerator v subscript 1 end subscript t over denominator t subscript 1 end subscript end fraction equals fraction numerator m v subscript 1 end subscript superscript 2 end superscript t over denominator t subscript 1 end subscript superscript 2 end superscript end fraction
    General
    physics-

    An aeroplane is flying in a horizontal direction with a velocity 600 k m h to the power of negative 1 end exponent at a height of 1960 m. when it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance A B

    From h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
    We have t subscript O B end subscript equals square root of fraction numerator 2 h subscript O A end subscript over denominator g end fraction end root
    equals square root of fraction numerator 2 cross times 1960 over denominator 9.8 end fraction end root equals 20 s
    Horizontal distance A B equals v t subscript O B end subscript
    equals open parentheses 600 cross times fraction numerator 5 over denominator 18 end fraction close parentheses left parenthesis 20 right parenthesis
    equals 3333.33 blank m equals 3.33 blank k m

    An aeroplane is flying in a horizontal direction with a velocity 600 k m h to the power of negative 1 end exponent at a height of 1960 m. when it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance A B

    physics-General
    From h equals fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent
    We have t subscript O B end subscript equals square root of fraction numerator 2 h subscript O A end subscript over denominator g end fraction end root
    equals square root of fraction numerator 2 cross times 1960 over denominator 9.8 end fraction end root equals 20 s
    Horizontal distance A B equals v t subscript O B end subscript
    equals open parentheses 600 cross times fraction numerator 5 over denominator 18 end fraction close parentheses left parenthesis 20 right parenthesis
    equals 3333.33 blank m equals 3.33 blank k m
    General
    physics-

    Three forces of magnitudes 6N, 6N and square root of 72 N at a corner of a cube along three sides as shown in figure. Resultant of these forces is

    The resultant of 5 N along O C and 5 N along O A is
    R equals square root of 6 to the power of 2 end exponent plus 6 to the power of 2 end exponent end root
    equals square root of 72N along O B
    The resultant of square root of 72 N along O B and square root of 72 N along O G is
    R to the power of ´ end exponent equals square root of 72 plus 72 end root equals 12N along O E.

    Three forces of magnitudes 6N, 6N and square root of 72 N at a corner of a cube along three sides as shown in figure. Resultant of these forces is

    physics-General
    The resultant of 5 N along O C and 5 N along O A is
    R equals square root of 6 to the power of 2 end exponent plus 6 to the power of 2 end exponent end root
    equals square root of 72N along O B
    The resultant of square root of 72 N along O B and square root of 72 N along O G is
    R to the power of ´ end exponent equals square root of 72 plus 72 end root equals 12N along O E.
    General
    maths-

    If the equation fraction numerator x to the power of 2 end exponent over denominator 9 minus k end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 minus k end fraction equals 1 represents an ellipse then

    If the equation fraction numerator x to the power of 2 end exponent over denominator 9 minus k end fraction plus fraction numerator y to the power of 2 end exponent over denominator 5 minus k end fraction equals 1 represents an ellipse then

    maths-General
    General
    physics-

    A particle of mass m moving with horizontal speed 6 blank m divided by s e c as shown in figure. If m less than less than M than for one dimensional elastic collision, the speed of lighter particle after collision will be


    v subscript 1 end subscript equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses u subscript 1 end subscript plus fraction numerator 2 m subscript 2 end subscript u subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
    Substituting m subscript 1 end subscript equals 0 comma blank v subscript 1 end subscript equals negative u subscript 1 end subscript plus 2 u subscript 2 end subscript
    rightwards double arrow v subscript 1 end subscript equals negative 6 plus 2 open parentheses 4 close parentheses equals 2 blank m divided by s
    i. e. the lighter particle will move in original direction with the speed of 2 blank m divided by s

    A particle of mass m moving with horizontal speed 6 blank m divided by s e c as shown in figure. If m less than less than M than for one dimensional elastic collision, the speed of lighter particle after collision will be

    physics-General

    v subscript 1 end subscript equals open parentheses fraction numerator m subscript 1 end subscript minus m subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction close parentheses u subscript 1 end subscript plus fraction numerator 2 m subscript 2 end subscript u subscript 2 end subscript over denominator m subscript 1 end subscript plus m subscript 2 end subscript end fraction
    Substituting m subscript 1 end subscript equals 0 comma blank v subscript 1 end subscript equals negative u subscript 1 end subscript plus 2 u subscript 2 end subscript
    rightwards double arrow v subscript 1 end subscript equals negative 6 plus 2 open parentheses 4 close parentheses equals 2 blank m divided by s
    i. e. the lighter particle will move in original direction with the speed of 2 blank m divided by s
    General
    physics-

    A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is

    By applying law of conservation of energy
    m g R equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent rightwards double arrow v equals square root of 2 R g end root

    A mass m slips along the wall of a semispherical surface of radius R. The velocity at the bottom of the surface is

    physics-General
    By applying law of conservation of energy
    m g R equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent rightwards double arrow v equals square root of 2 R g end root
    General
    physics-

    A 10 k g mass moves along x-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x equals 0 to x equals 8 blank c m

    Work done = Area covered in between force displacement curve and displacement axis
    = Mass cross timesArea covered in between acceleration-displacement curve and displacement axis
    equals 10 cross times fraction numerator 1 over denominator 2 end fraction open parentheses 8 cross times 10 to the power of negative 2 end exponent cross times 20 cross times 10 to the power of negative 2 end exponent close parentheses equals 8 cross times 10 to the power of negative 2 end exponent J

    A 10 k g mass moves along x-axis. Its acceleration as a function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x equals 0 to x equals 8 blank c m

    physics-General
    Work done = Area covered in between force displacement curve and displacement axis
    = Mass cross timesArea covered in between acceleration-displacement curve and displacement axis
    equals 10 cross times fraction numerator 1 over denominator 2 end fraction open parentheses 8 cross times 10 to the power of negative 2 end exponent cross times 20 cross times 10 to the power of negative 2 end exponent close parentheses equals 8 cross times 10 to the power of negative 2 end exponent J
    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

    At A comma blank v subscript A end subscript equals square root of g l end root
    At 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
    At A comma blank v subscript A end subscript equals square root of g l end root
    At 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 simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

    Vertical height equals h equals l cos invisible function application 30 degree
    Loss of potential energy equals m g h

    equals m g l cos invisible function application 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction m g l
    therefore Kinetic energy gained equals fraction numerator square root of 3 over denominator 2 end fraction m g l

    A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is

    physics-General
    Vertical height equals h equals l cos invisible function application 30 degree
    Loss of potential energy equals m g h

    equals m g l cos invisible function application 30 degree equals fraction numerator square root of 3 over denominator 2 end fraction m g l
    therefore Kinetic energy gained equals fraction numerator square root of 3 over denominator 2 end fraction m g l
    General
    physics-

    The potential energy of a particle varies with distance x as shown in the graph.
    The force acting on the particle is zero at

    F equals fraction numerator negative d U over denominator d x end fraction it is clear that slope of U minus x curve is zero at point B and C
    therefore F equals 0 for point B and C

    The potential energy of a particle varies with distance x as shown in the graph.
    The force acting on the particle is zero at

    physics-General
    F equals fraction numerator negative d U over denominator d x end fraction it is clear that slope of U minus x curve is zero at point B and C
    therefore F equals 0 for point B and C
    General
    maths-

    P(θ) and D space open parentheses theta plus pi over 2 close parentheses are the pts. on the ellipse b x to the power of 2 end exponent plus a to the power of 2 end exponent y to the power of 2 end exponent equals a to the power of 2 end exponent b to the power of 2 end exponent then C P to the power of 2 end exponent plus C D to the power of 2 end exponent equals

    P(θ) and D space open parentheses theta plus pi over 2 close parentheses are the pts. on the ellipse b x to the power of 2 end exponent plus a to the power of 2 end exponent y to the power of 2 end exponent equals a to the power of 2 end exponent b to the power of 2 end exponent then C P to the power of 2 end exponent plus C D to the power of 2 end exponent equals

    maths-General
    General
    physics-

    A ball of mass 0.2 blank k g rests on a vertical post of height 5 blank m. A bullet of mass 0.01 blank k g, travelling with a velocity V blank m divided by s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 blank mand the bullet at a distance of 100 blank mfrom the foot of the post. The initial velocity V of the bullet is

    R equals u square root of fraction numerator 2 h over denominator g end fraction end root rightwards double arrow 20 equals V subscript 1 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root and 100 equals V subscript 2 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root
    rightwards double arrow V subscript 1 end subscript equals 20 blank m divided by s comma blank V subscript 2 end subscript equals 100 blank m divided by s
    Applying momentum conservation just before and just after the collision open parentheses 0.01 close parentheses open parentheses V close parentheses equals open parentheses 0.2 close parentheses open parentheses 20 close parentheses plus left parenthesis 0.01 right parenthesis left parenthesis 100 right parenthesis
    V equals 500 blank m divided by s

    A ball of mass 0.2 blank k g rests on a vertical post of height 5 blank m. A bullet of mass 0.01 blank k g, travelling with a velocity V blank m divided by s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 blank mand the bullet at a distance of 100 blank mfrom the foot of the post. The initial velocity V of the bullet is

    physics-General
    R equals u square root of fraction numerator 2 h over denominator g end fraction end root rightwards double arrow 20 equals V subscript 1 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root and 100 equals V subscript 2 end subscript square root of fraction numerator 2 cross times 5 over denominator 10 end fraction end root
    rightwards double arrow V subscript 1 end subscript equals 20 blank m divided by s comma blank V subscript 2 end subscript equals 100 blank m divided by s
    Applying momentum conservation just before and just after the collision open parentheses 0.01 close parentheses open parentheses V close parentheses equals open parentheses 0.2 close parentheses open parentheses 20 close parentheses plus left parenthesis 0.01 right parenthesis left parenthesis 100 right parenthesis
    V equals 500 blank m divided by s
    General
    physics-

    The trajectory of a particle moving in vast maidan is as shown in the figure. The coordinates of a position A are open parentheses 0 , 2 close parentheses. The coordinates of another point at which the instantaneous velocity is same as the average velocity between the points are

    The trajectory of a particle moving in vast maidan is as shown in the figure. The coordinates of a position A are open parentheses 0 , 2 close parentheses. The coordinates of another point at which the instantaneous velocity is same as the average velocity between the points are

    physics-General
    General
    physics-

    A particle P is sliding down a frictionless hemispherical bowl. It passes the point Aat t equals 0. At this instant of time, the horizontal component of its velocity v. A bead Q of the same mass as P is ejected from A to t equals 0 along the horizontal string A B (see figure) with the speed v. Friction between the bead and the string may be neglected. Let t subscript p end subscript and t subscript Q end subscript be the respective time taken by P and Q to reach the point B. Then

    For particle P, motion between A and C will be an accelerated one while between C and B a retarded one. But in any case horizontal component of it’s velocity will be greater than or equal to v on the other hand in case of particle Q, it is always equal to v. Horizontal displacement of both the particles are equal, so t subscript P end subscript less than t subscript Q end subscript

    A particle P is sliding down a frictionless hemispherical bowl. It passes the point Aat t equals 0. At this instant of time, the horizontal component of its velocity v. A bead Q of the same mass as P is ejected from A to t equals 0 along the horizontal string A B (see figure) with the speed v. Friction between the bead and the string may be neglected. Let t subscript p end subscript and t subscript Q end subscript be the respective time taken by P and Q to reach the point B. Then

    physics-General
    For particle P, motion between A and C will be an accelerated one while between C and B a retarded one. But in any case horizontal component of it’s velocity will be greater than or equal to v on the other hand in case of particle Q, it is always equal to v. Horizontal displacement of both the particles are equal, so t subscript P end subscript less than t subscript Q end subscript