Question

A lift starts moving up with an acceleration of $5{\text{ms}}^{-2}$, and at the same instant a ball is dropped from a height of $1.25\text{m}$. The time taken by the ball to reach the floor of the lift is (nearly)
$(g=10{\text{ms}}^{-2})$

• A. $0.3$ second
• B. $0.2$ second
• C. $0.16$ second
• D. $0.4$ second

Answer 1

The correct option is D $0.4$ second

Taking upward as positive,
Initial velocity of the ball $=v_b=0$
Initial velocity of the lift $u_l=0$
Distance travelled by the lift w.r.t ball is given by
$S_{lb}=1.25\text{m}$
Acceleration of the lift $=a=5{\text{m/s}}^2$
Acceleration of the ball $=g=-10{\text{m/s}}^2$ (because $a$ and $g$ are in opposite direction)
Using second equation of motion, we can say
$S_{lb}=u_{lb}t+\frac{1}{2}{a}_{lb}{t}^2$
$\Rightarrow 1.25=0+\frac{1}{2}(5+10)t^2$
$\Rightarrow t=\sqrt{\frac{2\times 1.25}{5+10}}$
$\approx 0.4\text{s}$
Hence, the ball will take 0.4 seconds to touch the floor of the lift.