Suppose you drop a ball from a window 15 meters above the ground. The ball bounces to 60% 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.) 37.3 meters

B.) 378.6 meters

C.) 59.5 meters

D.) 63.4 meters

Redsox123  May 11, 2017

-15 +(30 + 18+ ....... )


-15+( sum of a GP with a=30, r=0.6 n=10)


\(-15+\frac{a(1-r^n)}{1-r}\\ =-15+\frac{30*(1-0.6^{10})}{1-0.6}\\ =-15+\frac{30*(1-0.6^{10})}{0.4}\\\)


-15+(30*(1-0.6^10))/0.4 = 59.5465 m


This is to the 10th bounce but it does not include the 10th bounce


Feel free to ask questions :)

Melody  May 11, 2017

9 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.