Question

Suppose a ball is dropped from a height of 6 ft. It bounces back up. But each time it bounces, it reaches only 7/10 of its previous height. What is the total distance the ball bounces (distance falling and going up) through the 5th bounce?

Answer 1

1st bounce 7/10 * 6 = 42/10 = 21/5
2nd bounce 7/10 * 21/5 = 147/50
3rd bounce 7/10 * 147/50 = 1029/500
4th bounce 7/10 * 1029/500 = 7203 / 5000
5th bounce 7/10 * 7203/5000 = 50421/50000 = 1.00842

Answer 2

Answer: 16.6386 ft.

Explanation:

Given: A a ball is dropped from a height of 6 ft.

Each time it bounces, it reaches only 7/10 of its previous height.

i.e. after 1st bounce the height reached by the ball is given by :-

`6*(7)/(10)`

After 2nd bounce the height reached by the ball is given by :-

`6*(7)/(10)*(7)/(10)=6*((7)/(10))^2`

Thus, it is following a geometric progression with common ratio (r)=7/10=0.7 and the first term = 6

The sum of geometric progression of first n terms is given by ;-

`(a(1-r^n))/(1-r)`

Now, the total distance the ball bounces (distance falling and going up) through the 5th bounce is given by:-

`(6(1-(0.7)^5))/(1-0.7)=16.6386`