Question

A sharp stone of mass $2$ kg falls from a height of $10$ m on sand and buries into the sand. It comes to rest in a time of $0.029$ second. The depth through which it buries into sand is

• A. 0.2 m
• B. 0.15 m
• C. 0.25 m
• D. 0.30 m

Answer 1

Answer: (A)

0.2 m

Velocity attained just before contact:

$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 10}=\sqrt{196}=14m/s$
$u_1=v$
$v_1=u_1+at$
$0=14+a\left(0.029\right)$
$a=-\left(\frac{14}{0.029}\right)$

$s=ut+\frac{1}{2}a{t}^2=14\left(0.029\right)+\frac{1}{2}\left(\frac{-14}{0.029}\right)(0.029{)}^2$
$s=0.029(14+\frac{1}{2}(-14\left)\right)$

$=0.029\times 7$
$=0.203m$
$=0.2m$