Q165 / 180 · KCET 2013 · Physics
In a lift moving up with an acceleration of $5 \mathrm{~ms}^{-2}$, a ball is dropped from a height of $1.25 \mathrm{~m}$. The time taken by the ball to reach the floor of the lift is ... (nearly) $\left(g=10 \mathrm{~ms}^{-2}\right)$
- A$0.3 \mathrm{~s}$
- B$0.2 \mathrm{~s}$
- C$0.16 \mathrm{~s}$
- Correct answerD$0.4 \mathrm{~s}$
Solution
Distance travelled by ball during fall $=1.25 \mathrm{~m}$
By using relation $s=u t+\frac{1}{2} a t^{2}$
Here, $u=0, \quad s=1.25 \mathrm{~m}$,
$\begin{aligned}
a &=(g+a)=(10+5)=15 \mathrm{~m} / \mathrm{s}^{2} \\
\Rightarrow \quad 1.25 &=\frac{1}{2} \times 15 \times t^{2}, \Rightarrow t^{2}=\frac{2 \times 1.25}{15} \\
t^{2} &=0.16 \quad \Rightarrow t=0.4 \mathrm{~s}
\end{aligned}$
Hence, the correct answer is option: 'D'.