Question

If the energy of a ball falling from a
height of 10 metres is reduced by
40%, how high will it rebound?
(Ans: 6 m)​

Answer 1

Answer:

Explanation:

As the GPE of a falling body is given by GPE = mgh, then if 40% of the total energy is turned into other forms, only 60% will remain as GPE when it bounces to its maximum height, and its rebound height will be 60% of its original height. If it was droppped from an original height of 10m, it will rebound to 6m.

Answer 2

Answer:

here you go .....

Explanation:

The Velocity of the ball just before striking the ground = $\sqrt{2gh}$

height = 10m, g = 10m/s²

velocity = 10$\sqrt{2}$m/s

kinetic energy = 1/2 mv²

kinetic energy = 1/2 m(200)

after hitting the ground the energy is reduced by 40percent

So final kinetic energy = 60percent of initial K.E

1/2mv$v_{2}$² = 1/2m(200) x 60/100

$v_{2}$² = 120

height to which it will rebound = v^2 / 2g

height = 120 / 20

height = 6m

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