geometrical sequence ball problem
A ball is dropped from a vertical height of 40 ft. Each time the ball bounces, it loses energy and only returns to a height that is 80% of the previous height. This process continues till the ball comes to rest. What is the total distance travelled by the ball by the time it comes to rest?
Considering $a_0=40$ (thus $a_1=32$), and using the infinite geometric series formula $a_1/(1-r)$, I got $32/(1/5) = 160$ft. However, the answer states it is 360ft. How can I get this answer?
$\endgroup$ 12 Answers
$\begingroup$You're only adding up the sum of the high points of all the ball's bounces. That's the total distance that the ball travels upwards (first upwards to $32$ feet on the first bounce, then upwards $25.6$ feet on the second bounce, and so on). But the ball also travels downwards to the ground prior to each bounce. That is, the total distance travelled by the ball is actually $40$ feet down plus $32$ feet up plus $32$ feet down plus $25.6$ feet up plus $25.6$ feet down and so on.
So to get the total distance travelled, you need to double what you found (after each bounce up, it falls the same distance back down), and add one more term of $40$ for the initial drop. That is, the total distance travelled is $2\cdot 160+40=360$ feet.
$\endgroup$ $\begingroup$$$ s = \left(40 + 2\cdot \frac{32}{1 -0,8}\right) ft = (40 +320)ft = 360 ft.$$
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