Question

A stone dropped from the top of a building taken 5 s to reach the ground. If it is stopped momentarily 4 s after it is dropped and then released again, how much time would it take from the moment it is released again to reach the ground? $(Takeg=10{ms}^{-2})$

• A. 1 s
• B. 3 s
• C. 4 s
• D. 5 s

Answer 1

Answer: (B)

3 s

First Case-

Let the distance covered by the stone in 5 s is $h$.

Using $s=ut+\frac{1}{2}g{t}^2$ $\Rightarrow h=0\times 5+\frac{1}{2}\times 10\times 5^2$ ( $\because u=0$)

$\Rightarrow h=125m$,

Second Case-

Let the distance covered by the stone in 4 s is $h_1$.

Using $s=ut+\frac{1}{2}g{t}^2$ $\Rightarrow h=0\times 4+\frac{1}{2}\times 10\times 4^2$ ( $\because u=0$)

$\Rightarrow h_1=80m$

At this moment stone is stopped and then released,

Let it take $t$ seconds to cover the remaining distance.

Remaining distance = $h-h_1$= 125-80=45 $m$,

Using $s=ut+\frac{1}{2}g{t}^2$ $\Rightarrow 45=0\times t+\frac{1}{2}\times 10\times t^2$ ( $\because u=0$)

$\Rightarrow t=3s$