Question

In a lift moving up with an acceleration of $5{\text{ms}}^{-2}$, 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 approximately $(g=10{\text{ms}}^{-2})$

• 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 $y$-axis as positive
Relative velocity of the lift with respect to ball $u_{lb}=0\text{m/s}$
Distance travelled by the lift w.r.t ball $=S_{lb}=1.25\text{m}$
Acceleration of the lift $=a=5{\text{m/s}}^2$
Acceleration due to gravity $=g=-10{\text{m/s}}^2$ (because $a$ and $g$ are in opposite direction)
Acceleration of the lift w.r.t ball $=a_{lb}=a–(-g)=5–(-10)=15{\text{m/s}}^2$
Using second equation of motion, we get
$S_{lb}=u_{lb}.t+\frac{1}{2}{a}_{lb}{t}^2$
$\Rightarrow 1.25=\frac{1}{2}\times 15\times t^2$
$t=\sqrt{\frac{2\times 1.25\text{m}}{(5+10){\text{ms}}^{-2}}}\approx 0.4\text{s}$
Hence, the ball will take nearly $0.4$ seconds to touch the floor of the lift.