Work Energy And Power Question 132

Question: A ball is projected vertically down with an initial velocity from a height of 20 m onto a horizontal floor. During the impact it loses 50% of its energy and rebounds to the same height. The initial velocity of its projection is [EAMCET (Engg.) 2000]

Options:

A) $ 20m{s^{-1}} $

B) $ 15m{s^{-1}} $

C) $ 10m{s^{-1}} $

D) $ 5m{s^{-1}} $

Show Answer

Answer:

Correct Answer: A

Solution:

Let ball is projected vertically downward with velocity v from height h .

Total energy at point $ A=\frac{1}{2}mv^{2}+mgh $ .

During collision loss of energy is 50% and the ball rises up to same height.

It means it possess only potential energy at same level. 50% $ ( \frac{1}{2}mv^{2}+mgh )=mgh $

$ \frac{1}{2}( \frac{1}{2}mv^{2}+mgh )=mgh $

$ v=\sqrt{2gh}=\sqrt{2\times 10\times 20} $ \ $ v=20m/s $



NCERT Chapter Video Solution

Dual Pane