A Bouncing Ball Made The Problem

When the ball falls down, it travels 8m,6m.9/2m,...........

since the distances traveled by the ball while bouncing is in G.P.

6m,9/2m,27/8m,81/32m

We can find the distances traveled by the the ball by using the formula

sum of the infinite terms in G.P.(S) =a/1-r, where a=1st term in G.P. ,r=common ratio between the two consecutive terms

Since the ball travels two times the distance it traveled while one bouncing

therefore the distance is 2a/1-r + 8

by substituting the values, we get

Total distance traveled by the ball=2(6)/1-3/4 + 8

=12/1/4 + 8

=48 + 8

=56

therefore the total distance traveled by the ball before it comes to rest is 56

The total distance $d$ the ball travels is: $d = 8 \text{ m} + 2\cdot6 \text{ m} + 2\cdot\frac{9}{2} \text{ m} + \cdots$ $= 2\sum_{n=1}^\infty 8\left(\frac{3}{4}\right)^{n-1} \text{ m} - 8 \text{ m}$ $= 2\left(\frac{8}{1-\frac{3}{4}}\right) \text{ m} - 8 \text{ m}$ $= \boxed{56} \text{ m}$

coefficient of restitution is e= $\sqrt { \frac { h1 }{ h } }$ .

where, h1=height after 1 collision = 24/4 = 6m.

h = original height = 8m.

thus, e^2 = 0.75.

total distance travelled before the ball stops is $h(\frac { 1+{ e }^{ 2 } }{ 1-{ e }^{ 2 } } )$ .

= $8(\frac { 1+{ 0.75 } }{ 1-{ 0.75 } } )$ = 8 x 7 = $\boxed{56}$