Question

A body falling from a vertical height of $10m$ pierces through a distance of $1m$ in sand. It faces an average retardation in sand equal to $(g=$ acceleration due to gravity)

• A. g
• B. 9 g
• C. 100 g
• D. 1000 g

Answer 1

Answer: (B)

If the ball is dropped then $x=0$, the velocity with which it will hit the sand will be given by
$v^2-u^2=2(-g)(-9)$
$v^2-0=18g$
$v^2=18g$ ...(i)
Now on striking sand, the body penetrates into sand for $1m$ and comes to rest.
So, $v\to$ initial for sand and final velocity $=0$
$v^{{}^{\prime }2}-v^2=2(a)\times (-1)$
$\Rightarrow -18g=-2a$
$\Rightarrow a=9g$