Question

3.A freely falling body strikes a horizontal surface and raises to a

height 1/64 of the height from which it was dropped. If the

velocity with which it hits the floor is v, the change in the

magnitude of velocity due to impact with the floor is

a) 7v/8 b) 5v/8 c) 3v/8 d) 9v/8

4. In the previous question, magnitude of the change in the of

velocity of the body due to impact is

a) 7v/8 b) 5v/8 c) 3v/8 d) 9v/8​

Answer 1

Answer:

a)7v/8

Explanation:

First of all

Change in Magnitude=Magnitude of Change

So the answer to 3 & 4 should be same

v1=(2gh)^1/2{I use ^1/2 to denote square root}

v2=(2gh/64)^1/2=(2gh)^1/2/8

v1-v2=(2gh)^1/2 - (2gh)^1/2/8

        =(8(2gh)^1/2-(2gh)^1/2)/8

        =(7(2gh)^1/2)/8

As v1=(2gh)^1/2,

v1-v2=7v1/8 or 7v/8