A ball dropped from a height of 10 feet bounces back of that distance. With each successive bounce, the ball continues to reach of its previous height. What is the total vertical distance (both up and down) traveled by the ball when it stops bouncing? (Hint: Add the total distance the ball falls to the total distance it rises.)

Question

A ball dropped from a height of 10 feet bounces back  of that distance.  With each successive bounce, the ball continues to reach  of its previous height.  What is the total vertical distance (both up and down) traveled by the ball when it stops bouncing?  (Hint:  Add the total distance the ball falls to the total distance it rises.)

anonymous · Answer

This ball travels according to $$\sum_{n = 1}^{\infty} 2*10(1/2)^n$$ because it bounces half as high each time, but travels both the up and down distance. This is equal to 20/(1-(1/2) = 40.

However, the first term here is the bounce up to height 5. We need to add in that initial drop from 10. 

40 + 10 = 50.