A ball is dropped vertically from height \(h\) and bounces elastically on the floor (see figure). Which of the following plots best depicts the acceleration of the ball as a function of time?
                      

1. 2.
3. 4.
Subtopic:  Uniformly Accelerated Motion |
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A particle projected vertically under gravity passes a certain level on the way up at a time \(T_1\) and on the way down at a time \(T_2\) – after it was projected. The speed of projection is:

1. \(\dfrac{1}{2} g\left(T_{1}+T_{2}\right)\)

2. \(\dfrac{1}{2} g\left(T_{1}-T_{2}\right)\)

3. \(g \sqrt{T_{1} T_{2}}\)

4. \(\dfrac{1}{2} g \dfrac{T_{1} T_{2}}{T_{1}+T_{2}}\)

Subtopic:  Uniformly Accelerated Motion |
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A boy throws a ball straight up the side of a building and receives it after \(4~\text s.\) On the other hand, if he throws it so that it strikes a ledge on its way up, it returns to him after \(3~\text s.\) The ledge is at a distance \(d\) below the highest point, where \(d=? \) (take acceleration due to gravity, \(g=10~\text{ms}^{-2})\)

1. \(5~\text m\) 2. \(2.5~\text m\)
3. \(1.25~\text m\) 4. \(10~\text m\)
Subtopic:  Uniformly Accelerated Motion |
Level 3: 35%-60%
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A man driving a scooter at \(15~\text{m/s}\) brakes at the rate of \(2~\text{m/s}^2\). His speed, after \(10~\text{s}\) after the application of brakes will be:
1. \(5~\text{m/s}\)
2. \(-5~\text{m/s}\)
3. \(0~\text{m/s}\)
4. \(10~\text{m/s}\)

Subtopic:  Uniformly Accelerated Motion |
Level 3: 35%-60%
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A ball is thrown vertically upwards with a velocity of \(19.6~\text{ms}^{-1}\) from the top of a tower. The ball strikes the ground after \(6~\text s.\) The height from the ground up to which the ball can rise will be \(\left ( \dfrac{k}{5} \right )~\text {m}.\)The value of \(k\) is: 
(use \(g=9.8~\text{ms}^{-2})\)
1. \(392 \)
2. \(360\)
3. \(315\)
4. \(420\)
Subtopic:  Uniformly Accelerated Motion |
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Two particles \(A\) & \(B\) start moving from the same point with initial velocities and accelerations:
Particles\(\rightarrow\) \(A\) \(B\)
initial velocity \(-\vec u\) \(\vec u\)
acceleration \(\vec a\) zero
The vector \(\overrightarrow{AB}\) is given by:
1. \(2\vec ut+{\large\frac12}\vec at^2\) 2. \(2\vec ut-{\large\frac12}\vec at^2\)
3. \({\large\frac12}\vec at^2\) 4. \(-{\large\frac12}\vec at^2\)
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A particle is projected vertically upwards with a speed \(u\) and moves under the force of gravity. The distance travelled by the particle during its entire motion (until it returns) is \(d_1.\) If the force of gravity were to be switched off, and the particle travelled for the same length of time, then the distance travelled is \(d_2.\) Then, 
1. \(d_2 = d_1\)
2. \(d_2 = 2d_1\)
3. \(d_2 = 3d_1\) 
4. \(d_2 = 4 d_1\)
Subtopic:  Uniformly Accelerated Motion |
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A small particle slides up and down along a smooth path \(ABC~(\angle B=90^{\circ}), \) under the action of gravity, coming to a stop briefly at the highest points \((A,C)\) of the path \(ABC.\) The path is rounded at \(B \) to facilitate the back and forth motion. The time taken for the particle to go from \(A\) to \(C\) is
      
 
1. \(\sqrt{\dfrac{2h}{g}}(\sin\theta+\cos\theta) \)
2. \(\sqrt{\dfrac{2h}{g}}(\sin^2\theta+\cos^2\theta) \)
3. \(\sqrt{\dfrac{2h}{g}}\left(\dfrac1{\sin\theta}+\dfrac1{\cos\theta}\right) \)
4. \(\sqrt{\dfrac{2h}{g}}\left(\dfrac1{\sin^2\theta\cdot \cos^2\theta}\right) \)
Subtopic:  Uniformly Accelerated Motion |
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A man \((A)\) has to throw a ball vertically up to a partner \((B)\) who is standing up, above his level by \(15~\text{m}.\) The \((B)\) partner can catch the ball only when it comes downwards with a maximum speed of \(10~\text{m/s}\)
(take acceleration due to gravity as \(10~\text{m/s}^{2}\))
The minimum and maximum speeds of the throw are: (nearly)
1. \(10~\text{m/s}~\text{and}~20~\text{m/s}\)
2. \(10~\text{m/s}~\text{and}~30~\text{m/s}\)
3. \(20~\text{m/s}~\text{and}~20\sqrt{3}~\text{m/s}\)
4. \(10\sqrt{3}~\text{m/s}~\text{and}~20~\text{m/s}\)

Subtopic:  Uniformly Accelerated Motion |
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