Question

A stone weighing $3\text{kg}$ falls from the top of a tower $100\text{m}$ high and buries itself $2\text{m}$ deep in sand. Find the time (in seconds) of penetration $(g=10{\text{m/s}}^2)$

• A. $\sqrt{5}$
• B. $2\sqrt{5}$
• C. $25\sqrt{5}$
• D. $\frac{1}{5\sqrt{5}}$

Answer 1

The correct option is D $\frac{1}{5\sqrt{5}}$


Let $C$ be the point deep inside the sand where the stone buries itself.
For $AB:$
Initial velocity $u=0$
$S=100\text{m},a=g=10{\text{m/s}}^2$
By using third equation of motion
$v_B^2-0=2gS$
$\Rightarrow v_B=20\sqrt{5}\text{m/s}$
For $BC:$
$v_c=0$
and $S_{BC}=2\text{m}$
By using third equation of motion
$v_C^2-v_B^2=-2a{S}_{BC}$
$0-2000=-2a\times 2$
$\Rightarrow a=500{\text{m/s}}^2$
Now, by first equation of motion,
$v_C=v_B-at$
$0=20\sqrt{5}-500t$
$\Rightarrow t=\frac{20\sqrt{5}}{500}=\frac{\sqrt{5}}{25}=\frac{1}{5\sqrt{5}}\text{s}$