Question

A boy is standing on a lift moving up with uniform speed of $9.8\, m/s$. He throws a ball up with speed of $98\, m/s$ w.r.t. lift. Assuming the lift to be opened from the top find the time taken by the ball to return to his hands.
A. 10 sec
B. 20 sec
C. 25 sec
D. 40 sec

Answer 1

The initial velocity of the ball will be the sum of velocity of throw and velocity of lift. In time t, the lift also has moved an upward distance and therefore, the displacement of the ball can be found using the kinematic equation. Use the kinematic equation to express the displacement of the ball and find the time taken by the ball to return to the boy.

Formula used:
$s = ut - \dfrac{1}{2}g{t^2}$
Here, s is the displacement, u is the initial velocity, g is the acceleration due to gravity and t is the time.

Complete step by step answer:
We have given that the boy throws a ball vertically upward with velocity 98 m/s with respect to lift and the lift is also moving up with the velocity 9.8 m/s. Therefore, the initial velocity of the ball will be the sum of velocity of throw and velocity of lift that is 107.8 m/s. The displacement s of the ball is equal to the vertical distance moved by the lift.

We can use the kinematic equation to express the displacement of the ball as follows,
$s = ut - \dfrac{1}{2}g{t^2}$
Here, s is the displacement, u is the initial velocity, g is the acceleration due to gravity and t is the time.

In time t, the lift also has moved an upward distance h. Let’s express the distance moved by the lift in time t as follows,
$s = u't$
Here, $u'$ is the velocity of the lift.
Therefore, the above equation becomes,
$u't = ut - \dfrac{1}{2}g{t^2}$
Substituting 9.8 m/s for $u'$, 107.8 m/s for u and $9.8\,{\text{m/}}{{\text{s}}^2}$for g in the above equation, we get,
$9.8t = \left( {107.8} \right)t - \dfrac{1}{2}\left( {9.8} \right){t^2}$
$\Rightarrow 98t = 4.9{t^2}$
$\therefore t = 20\,{\text{s}}$

Thus, the ball will return to the boy in time 20 sec. So, the correct answer is option B.

Note: Students often forget that the lift also moves in the same direction as the ball and the distance to be travelled for the ball to return to the boy reduces by some amount. Therefore, always note that the displacement of the ball is not zero because it is not thrown from the stationary lift. Also, if the velocity of the ball is given with respect to the boy, the initial velocity of the ball will be equal to the initial velocity of the ball thrown from the stationary lift.