Question

A $5kg$ stone falls from a height of $1000m$ and penetrates $2m$ in a layer of sand. The time of penetration is

• A. 14.285 s
• B. 0.0285 s
• C. 7.146 s
• D. 0.285 s

Answer 1

Answer: (B)

0.0285 s

Speed of the stone just before touching the sand is
$v=\sqrt{2gh}=\sqrt{2\times 9.81\times 1000}=140.07m/s$
Deceleration of the stone due to sand is
$v^2-u^2=-2as$
$0-{140.07}^2=-2\left(a\right)\left(2\right)$
$a=4905m/s^2$
We have initial speed and the deceleration, from
$v=u-at$
$u=at$
$t=\frac{u}{a}=\frac{140.07}{4905}=0.0285s$
Option B.