SATHEE: Work Energy And Power Question 132

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) $20 m s^{- 1}$

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

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

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

Show Answer

Answer:

Correct Answer: A

Solution:

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

Total energy at point $A = \frac{1}{2} m v^{2} + m g h$ .

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% $(\frac{1}{2} m v^{2} + m g h) = m g h$

$\frac{1}{2} (\frac{1}{2} m v^{2} + m g h) = m g h$

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

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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:

Answer:

Solution:

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