Question

A ball was dropped from a height of 60m. Each time it hit the ground, it bounced up 1/2 (half) of the height that it dropped. How far had the ball travelled when it hit the ground for the fifth time?

Answer 1

Answer:

The ball traveled 116.25 m when it hit the ground for the fifth term

Step-by-step explanation:

This is a geometric progression exercise and what we are asked to look for is the sum of a GP.

The ball was dropped from a height of 60 m. This means that the initial height of the ball is 60 m.

First value, a = 60

Each time it hit the ground, it bounced up 1/2 (half) of the height that it dropped.

This is the common ratio, r = 1/2 = 0.5

The number of terms it hits the ground is the number of terms in the GP.

number of terms, n = 5

The distance traveled by the ball when it hit the ground for the fifth term will be modeled by the equation:

$S_n = \frac{a(1 - r^n) }{1 - r} \\S_5 = \frac{60(1 - 0.5^5) }{1 - 0.5}\\S_5 = \frac{58.125}{0.5} \\S_5 = 116.25 m$