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

s = 10/ 100

s = 0.1 m

velocity = root x 2 x g x h

= root 2 x 10 x 5

= root 100

= 10 m/ s

v^2 - u^2 = 2x a x s

100 = 2 x a x 0.1

a = 500 m/s^2

Answer 2

Answer:

Explanation:

v2 = u2 + 2as

v2 = 2*10*5

v= 10m/s

Again,

v2=u2+2as

a= -u2/2s

a = 500 m/s2

But,

This is net acceleration on the ball

i.e acceleration of sand + acceleration due to gravity

Thus deceleration offered by sand = (500-10) = 490 m/s2