Respuesta :
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:
[tex]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[/tex]
