Aball is dropped from a height of 16 feet. each time it is drops, it rebounds 80% of the height from which it is falling fine the total distance traveled in 15 bounces

A sequence of height reached by the ball is a geometric sequence or progression

The total distance traveled in 15 bounces is approximately 77.185 feet

Reason:

The given parameters are;

Height from which the ball dropped = 16 feet

Height to which the ball rebounds to = 80% of the previous height

The total distance traveled in 15 bounces = Required

Solution;

The height traveled at each bounce is given by a geometric progression, of the form, aₙ = a·rⁿ

Where;

The first term, a = 16 feet

Common ratio, r = 0.8

The sum of n, terms of a geometric progression, Sₙ, is given as follows;

The total distance traveled in 15 bounces, Sₙ ≈ 77.185 feet

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Notice total distance is comprised of both positive movements and negative movements, and both sequences are geometric (exponential) sequences...

The sum of a geometric sequence is:

s(n)=a(1-r^n)/(1-r), a=initial term, r=common ratio, n=term number

So for the "dropping" distances you have the sum...

s(n)=16(1-.8^n)/0.2=80(1-.8^n)

s(15)=80(1-.8^15)

And the "rising" distances you the first term is .8(16)=12.8 and n=14 so

s(14)=12.8(1-.8^14)/(.2)

s(14)=64(1-.8^14)

So the total distance traveled is:

80(1-.8^15)+64(1-.8^14)

138.37050046578688

Total distance is approximately 138.37 ft

step-by-step explanation:

yes it is1. move over 2 points on the x-axis2. move over 2 points on the y-axis