A ball is dropped from height h on a horizontal floor. If it loses 60% of its energy on hitting the floor then upto which it will rise after first rebounce is?

60 percent of the energy is lost. Then use energy conservation as follows:

mgh=mgh1+(60/100)mgh

Here, m is the mass of the ball, h is the height from where the ball is dropped, h1 is the required height and g is the gravitational acceleration.

On further solving the above equation, we get,

mgh=mgh1 + (60100)mgh

h = h1 + 0.6hh1 = h(1−0.6)

h1 = 0.4h

On simplifying the above relation as,

h1 = 0.4hh1 = 2/5h

Therefore, the ball will rebound upto the height of 2/5 times the given height.

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