Question

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

Answer 1

Before the ball hits the ground,
u = 0
a = 10 $m/s^2$ (in the downward direction)
s = 5 m
From the third equation of motion, $v^2=u^2+2as$
$\Rightarrow v^2=2\times 10\times 5$
$\Rightarrow v=10m/s$
Now, the ball enters the sand with this initial velocity of 10 m/s and is uniformly retarded to rest.
u = 10 m/s
v = 0
s = 10 cm = 0.1 m

Again applying the third equation of motion, $v^2=u^2+2as$
$0={10}^2+2a\times 0.1$
$\Rightarrow a=-500m/s^2$
So, the retardation is $500m/s^2$