Question

A particle is thrown up 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 $g$ (where $g$ is the 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 $\frac{T}{2}$

For the first throw, time of flight $T=\frac{2V_0}{g}$
For the second throw, the velocity of particle is $V_0$ w.r.t 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 $a_{rel}=-g-g=-2g$
Using second equation of motion,
$S_{rel}=u_{rel}t+\frac{a_{rel}{t}^2}{2}$
$\Rightarrow 0=V_{0}t-\frac{2g{t}^2}{2}$
Solving this we get $t=0$ and $t=\frac{V_0}{g}$
$\because t=0$ is not possible
$\Rightarrow t=\frac{V_0}{g}=\frac{T}{2}$