Question

A ball is dropped from a height of $5m$ onto a sandy floor and penetrates the sand up to $10cm$ before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.

Answer 1

Consider the motion of ball from $A$ to $B$.

$B\to$ just above the sand ( just to penetrate)

$u=0,a=9.8m/s^2,s=5m$

The distance covered by the ball is:

$S=ut+\frac{1}{2}a{t}^2$

$\Rightarrow 5=0+\frac{1}{2}\left(9.8\right)t^2$

$\Rightarrow t^2=\frac{5}{4.9}=1.02$

$\Rightarrow t=1.01$.

$\therefore$ velocity at $B,v=u+at=9.8\times 1.01(u=0)=9.89m/s$.

From motion of ball in sand

$u_1=9.89m/s,v_1=0,a=?,s=10cm=0.1m$

$a=\frac{v_1^2-u_1^2}{2s}=\frac{0-(9.89{)}^2}{2\times 0.1}=-490m/s^2$

The retardation in sand is $490m/s^2$.