Answer:
[tex]s = \sum^5_1 {(2h)(\frac{2}{3})^i},[/tex]
Step-by-step explanation:
The initial height of the ball is h
After the first bounce the height is [tex]\frac{2}{3}h[/tex]
After the second bounce the height is [tex]\frac{2}{3}(\frac{2}{3})h[/tex]
After the i-th rebound the height is [tex](\frac{2}{3}) ^ i[/tex]
Then, distance s traveled by the ball is the sum of the heights reached between the first and fifth bounces.
[tex]s = 2 [\frac{2}{3}h + \frac{2}{3}(\frac{2}{3})h +, ..., + (\frac{2}{3}) ^ n h][/tex]
The equation is multiplied by 2 because the distance the ball travels when it goes up is the same as it travels down.
Finally the distance is a geometric series as shown:
[tex]s = \sum^5_1 {(2h)(\frac{2}{3})^i},[/tex]
