Suppose that you drop a ball from a window 45 meters above the ground. The ball bounces to 65% of its previous height with each bounce. What is the total number of meters (rounded to the nearest tenth) the ball travels between the time it is dropped and the 10th bounce?

A)208.7 meters

B)162.3 meters

C)199.5 meters

D)203.9 meters

Guest Dec 21, 2017

#1**+1 **

We can use this to find the total distance, D

2 * 45 [ 1 - .65^10] / [ 1 - .65] - 45 ≈ 208.7 m

CPhill
Dec 21, 2017

#2**+1 **

It falls 45 m initially then

First bounce (there is an up and a down so multiply by 2) = 2 x 45(.65)

+

Second bounce = 2 x 45(.65)(.65)

+

third bounce = 2 x 45 (.65)(.65)(.65)

+

.

.

.

ninth bounce

= 45 + 2 x 45 x\(\sum_{1}^{9}\).65^n = 45 + 2 x 45 (1.818677876) = 208.681 m (I think)

ElectricPavlov
Dec 21, 2017