Question

An open lift is coming down from the top of a building at a constant speed $v=10m/s$. A boy standing on the lift throws a stone vertically upwards at a speed $30m/s$ w.r.t. himself. The time after which he will catch the stone is:

• A. $4sec$
• B. $6sec$
• C. $8sec$
• D. $10sec$

Answer 1

Answer: (B)

$6sec$

The initial velocity of the stone w.r.t to the boy is 30 m/s. The lift is also coming down with a constant velocity 10 m/s. The time of flight of the stonedoesn not depend on the lift's speed. When the boy catches the stone totally depends on the speed of the ball w.r.t to the boy.

So, the time of flight of the ball is $t^{\prime }$ (let).

Now, $v=u-gt$

$or,0=30-10t$

$or,t=3$

so, Time of flight is $t^{\prime }=2t=2\times 3s=6s$

The correct answer is $OptionB$