Question

A body of mass $m$ falls from rest through a height $h$ under gravitation acceleration $g$ and is then brought to rest by penetrating through a depth $d$ into some sand. The average deceleration of the body during penetration into sand is

• A. $\frac{g{h}^2}{d^2}$
• B. $\frac{g{h}^2}{2d^2}$
• C. $\frac{gh}{d}$
• D. $\frac{gd}{h}$

Answer 1

Answer: (C)

$\frac{gh}{d}$

Here, $AB=h$ and $BC=d$
Velocity at point $B$,
$v^2=u^2+2gh$
$v^2=0+2gh$ $[\because u=0]$
$v^2=2gh$ ...(i)
Since, the body comes to rest at point $C$.
Then, $v=0$
Now, $v^2=u^2-2ad$ [where, $a$ is deceleration]
$\Rightarrow 0=u^2-2ad$
$\Rightarrow 2ad=u^2$
$\Rightarrow 2ad=2gh$ $[\because atpointB,u^2=2gh]$
$\Rightarrow a=\frac{gh}{d}$.