A ball is dropped from a height of 5.0 m. On each upward bounce the ball returns to $$\frac { 4 } { 5 }$$ of its previous height. Find the total vertical distance the ball travels before coming to rest.
A) 25.0 m
B) 50.0 m
C) 46.0 m
D) 45.0 m

Question

Quizplus · Accepted Answer

The total vertical distance includes the initial drop plus the sum of the distances traveled on each bounce up and down. The distance traveled on each bounce decreases geometrically by a factor of $\frac{4}{5}$. The series is: $5 + 2(5 \times \frac{4}{5}) + 2(5 \times (\frac{4}{5})^2) + 2(5 \times (\frac{4}{5})^3) + \ldots$. This is a geometric series with the first term $a = 5$ and common ratio $r = \frac{4}{5}$, where the factor of 2 accounts for the upward and downward journey except for the first drop. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$, so here $S = \frac{5}{1 - \frac{4}{5}} = \frac{5}{\frac{1}{5}} = 25$. Since the first term (the initial drop) is not doubled, we add another $5 \times 2$ for the subsequent bounces, giving $25 + 20 = 45$ meters.

A Ball Is Dropped from a Height of 5 45\frac { 4 } { 5 }54​

A Ball Is Dropped from a Height of 5 $\frac { 4 } { 5 }$