Question

The distance travelled by the ball during its flight is:

• A. $32.64m$
• B. $31.86m$
• C. $52m$
• D. $30.56m$

Answer 1

Answer: (D)

$30.56m$

Let initial point of floor of elevator be the reference point.

We have, $x=x_0+ut+\frac{1}{2}{at}^2$

$\therefore$ $x_{ball}=2+25t-\frac{1}{2}\times 10\times t^2\to \left(1\right)$

$x_{elevator}=0+10t+5t^2\to \left(2\right)$

The ball will hit the floor when

$x_{ball}=x_{elevator}$

On solving $\left(1\right)$ & $\left(2\right)$, $t=2.13s$

$\therefore$ Distance traveled by the ball during height is,

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

$=25\times 2.13-\frac{1}{2}\times 10\times {2.13}^2$

$=30.56m$