Question

A ball is dropped from an initial height and allowed to bounce repeatedly. On the first bounce (one up-and-down motion), the ball reaches a height of 32 inches. On each successive bounce, the ball reaches 75% of its previous height. What is the total vertical distance that the ball travels in 10 bounces? (Do not include the initial height from which the ball is dropped.)

Answer 1

S₁₀ = 241.5837 m

Step-by-step explanation:

If

h₁ = 2*32 in = 64 in

h₂ = 0.75*h₁ = (0.75)*64 in = 48 in

h₃ = 0.75*h₂ = 0.75*(0.75*h₁) = (0.75)²*h₁ = (0.75)²*64 in = 36 in

h₄ = 0.75*h₃ = 0.75*(0.75*h₁) = (0.75)³*h₁ = (0.75)³*64 in = 27 in

...

h₁₀ = 0.75*h₉ = (0.75)⁹*h₁ = (0.75)⁹*64 in = 4.8054 in

It is a geometric sequence (geometric progression) where the common ratio is

r = 0.75

Finding the sum of terms in a geometric progression is easily obtained by applying the formulas:

10th partial sum of a geometric sequence

S₁₀ = h₁*(1 - r¹⁰) / (1 - r)

⇒ S₁₀ = 64*(1 - 0.75¹⁰) / (1 - 0.75)

⇒ S₁₀ = 241.5837 m