Question

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

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

Answer 1

Answer: (D)

Distance travelled by ball during fall $=1.25\, m$
By using relation $s=u t+\frac{1}{2}\, a t^{2}$
Here, $u=0, s =1.25\, m$
$ a=(g+a)=(10+5)=15 \,m / s ^{2} $
$\Rightarrow 1.25 =\frac{1}{2} \times 15 \times t^{2}, $
$\Rightarrow t^{2}=\frac{2 \times 1.25}{15} $
$t^{2} =0.16 $
$ \Rightarrow t=0.4\, s$