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}} $
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Answer:
Correct Answer: A
Solution:
Let a ball be 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 half the original height.
It means it possesses only potential energy at same level. 50% $ ( \frac{1}{2}mv^{2}+mgh )=mgh $
$ \frac{1}{2}( \frac{1}{2}mv^{2}+mgh )=\frac{1}{2}mgh $
$ v=\sqrt{2gh}=\sqrt{2\times 9.81\times 20} $ \ $ v=20m/s $
BETA
