1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A ball is dropped from a hieght of 5m onto a Sandy floor and penetrates the sand upto 10 cm before coming to rest . Find the retardation of the ball in sand assuming it to be uniform.

Open in App
Solution

## Initial velocity(u) of the ball is 0 m/s. Let the final velocity of the ball when it just strikes the sand is v m/s. Applying Newton's laws,we get, v2-u2=2gs v2-0=2(10)(5) v2=100 v=10 m/s. Now, initial velocity(v) when the ball strikes the sand is 10 m/s and the final velocity v' when the ball stops inside the sand is 0 m/s. Here,displacement inside the sand is 10 cm or 0.1 m and the accn. of the ball inside the sand is a' m/s2. again,applying Newton's laws,we get, v'2-v2=2a's 0-100=2(-a)(0.1).........(here the negative sign indicates retardation of the ball,i.e.negative acc. of the ball.) -100=-0.2a a=500m/s2 So,the retardation of the ball is 500m/s2.

Suggest Corrections
10
Join BYJU'S Learning Program
Related Videos
Realistic Collisions
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program