Question

A very broad elevator is going up vertically with a constant acceleration of $2m/s^2$. At the instant when its velocity is $4m/s$ a ball is projected from the floor of the lift with a speed of $4m/s$ relative to the floor at an elevation of ${30}^{\circ }$. The time taken by the ball to return the floor is ($g$ = $10m/s^2$) from time of projection.

• A.
• B.
• C.
• D. 1 s

Answer 1

The correct option is B

$\frac{{}^{a}ball}{lift}=a_{ball}-a_{lift}$

=$-g\hat{j}-2\hat{j}$
=$-(g+2)\hat{j}$
=$-12\frac{m}{s^2}\hat{j}$

Now in the lifts frame, the ball is simply a projectile with

a = $-12\frac{m}{s^2}\hat{j}$ and not

a = $=-g\frac{m}{s^2}\hat{j}$

$\therefore$ Time of flight = $T$ = $\frac{2\mu sin\theta }{12}$

= $\frac{2\times 4\times sin30}{12}$

= $2\times 4\times \frac{1}{2}\times \frac{1}{12}$

= $\frac{1}{3}s$