Question

A boy throws up a ball in a stationary lift and the ball returns to his hands in $10s$. Now if the lift starts moving up at a speed of $5m{s}^{-1}$. The time taken for a ball thrown straight up to return to his hands is

Answer 1

Step 1: Given data:

The velocity of the lift is $v=5m/s.$

After throwing the ball it returns in $10s$ when the lift is stationary

The displacement of the ball is zero because the ball will return to its initial point.

Diagram:

Step 2: Formula used

We know from the formulae of kinematics that, the displacement of a body is given by the equation,

$S=ut+\frac{1}{2}a{t}^2$

Where, $S$ and $u$ are the displacement and initial velocity of the body, $a$ is the acceleration of the body and $t$ is the time.

Step 3: Calculation:

Let $u$ be the velocity projection of the ball.

The displacement of the body is zero (due to uniform velocity),$S=0$

Time taken, $t=10s$

$$\begin{gathered} S=ut-\frac{1}{2}g{t}^2 \\ \Rightarrow 0=\left(u\times 10\right)-\frac{1}{2}\times \left(9.8\right)\times {\left(10\right)}^2 \\ \Rightarrow 10u=50\times 9.8 \\ \Rightarrow u=\frac{50\times 9.8}{10}=49 \\ \Rightarrow u=49m{s}^{-1} \end{gathered}$$

So, the velocity of projection of the ball is $49m/s.$

When the lift is in motion

Due to uniform motion, the displacement of the lift is,

$$\begin{gathered} S_2=v{t}_2 \\ \Rightarrow S_2=5t_2...........\left(1\right) \end{gathered}$$

And the displacement of the ball when the lift is moving up,

$$\begin{gathered} S=(u+v)t_2-\frac{1}{2}g{t_2}^2 \\ \Rightarrow S=(49+5)t_2-\frac{1}{2}\times 9.8\times {t_2}^2 \\ \Rightarrow S=54t_2-4.9{t_2}^2..............\left(2\right) \end{gathered}$$

Due to uniform motion (velocity of the lift).

The displacement of the ball = displacement of the body. So on equating equations $\left(1\right)$ and $\left(2\right)$ we get,

$$\begin{gathered} S_2=S \\ \Rightarrow 5t_2=54t_2-4.9{t_2}^2 \\ \Rightarrow 4.9{t_2}^2=49t_2 \\ \Rightarrow 4.9t_2=49 \\ \Rightarrow t_2=\frac{49}{4.9}=10 \\ \Rightarrow t_2=10sec \end{gathered}$$

Therefore, after$10s$, the ball will return.