SOLUTION: A ball is thrown vertically upward to a height of 50 feet. Every time the ball strikes the ground, it rebounds 1/2 of the previous height. How far has the ball traveled up and down

Question 836838: A ball is thrown vertically upward to a height of 50 feet. Every time the ball strikes the ground, it rebounds 1/2 of the previous height. How far has the ball traveled up and down when it strikes the ground 6th time?
Answer by thejackal(72) (Show Source): You can put this solution on YOUR website!
This is an example of a geometric progression.
We know this because to arrive to the next term we must divide the height by half.
Since the maximum height reached is 50, then the total distance covered going up and down is 100. 100 is therefore our first term. the common ration in this case is 0.5 since 0.5 x 100 will give us the second term which is 50.
now using the formula for geometric progressions:
A(n) = A * R^(n-1) where A is the first term, n is the nth term we want and R=ratio. therefore
A(6) = 100 * 0.5^(6-1) = 3.125 ft up and down and 1.563 ft one way
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