Question

A ball is dropped from 64 M and rebounds three by fourth of the height again and again............ so find the distance travelled by the ball till the ball being at rest ..........the air is negligible.......... right answer should be mark as a brainliestant.......!!!​

Answer 1

Let the height above which the ball is released be H

This problem can be tackled using geometric progression.

Tn=ar^(n-1)

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

N

Sn= sigma ar^(n-1) = a(1- r^n)/(1-r)

1

To find the total distance travel one has to sum over up to n=3.n=3. But there is little subtle point here. For the first bounce (n=1n=1), the ball has only travel H and not 2H. For subsequent bounces (n=2,3,4,5......n=2,3,4,5......), the distance travel is 2×(3/4)n×H2×(3/4)n×H

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

However we have to subtract HH because up to the first bounce, the ball only travel HH instead of 2H2H

Therefore the total distance travel up to the Nth bounce is

Dn=

$2h \frac{1 - { (\frac{3}{4} )}^{n} }{1 - \frac{3}{4} }$

For N=3 one obtains

D=3.625H

D= 3.625 * 64=232