+0  
 
0
82
2
avatar

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
Sort: 

2+0 Answers

 #1
avatar+81063 
+1

We can use this to find the total distance, D

 

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

 

 

cool cool cool

CPhill  Dec 21, 2017
 #2
avatar+10859 
+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

19 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details