Question

A particle is thrown up with initial speed $V_0$ inside a stationary lift of sufficient height and the time of flight is $T$. Now it is thrown again with same initial speed $V_0$ with respect to the lift. At the time of second throw, lift is moving up with speed $V_0$ and uniform acceleration $g$ upwards (where $g$ is the acceleration due to gravity). The new time of flight will be

• A. $\frac{T}{4}$
• B. $\frac{T}{2}$
• C. $T$
• D. $2T$

Answer 1

The correct option is B $\frac{T}{2}$

Taking upward as positive for both cases,
For the first throw, time of flight $T=\frac{2V_0}{g}$
For the second throw, the velocity of particle is $V_0$ with respect to lift.
$V_{rel}=V_0$
Displacement of particle upwards would be same as displacement of lift upwards, so relative displacement has to be zero. i.e. $S_{rel}=0$
Relative acceleration of particle w.r.t lift is given by
$a_{rel}=-g-g=-2g$
Using second equation of motion,
$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$
$\Rightarrow 0=V_{0}t+\frac{1}{2}{a}_{rel}{t}^2\Rightarrow t=\frac{V_0}{g}=\frac{T}{2}$