Physics, asked by sairoshinimacharla, 8 months ago


A highly elastic ball is released from rest at a distance h above the ground and bounces up and down.
With each bounce a fraction f = 1/10 of its KE just before the bounce is lost. The time for which the ball will continue to bounce is nearly


Answers

Answered by shreewomenshostel201
1

Answer:

Let the height above which the ball is released be H

This problem can be tackled using geometric progression.

The nth term of a Geometric progression is given by the above, where n is the term index, a is the first term and the sum for such a progression up to the Nth term is

To find the total distance travel one has to sum over up to n=3. But there is little subtle point here. For the first bounce ( n=1 ), the ball has only travel H and not 2H. For subsequent bounces ( n=2,3,4,5...... ), the distance travel is 2×(3/4)n×H

a=2H..........r=3/4

However we have to subtract H because up to the first bounce, the ball only travel H instead of 2H

Therefore the total distance travel up to the Nth bounce is

For N=3 one obtains

D=3.625H

Explanation:

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