Question

A bouncing ball is dropped from an initial height of h meters above a flat surface. Each time the

ball hits the surface, it rebounds a distance r × h meters and it bounces indefinitely. Consider

the value h = 5 meters and r = 1/3. The total vertical distance (up and down) travelled (in meters)

by the ball is​

Answer 1

Answer:

This is actually a fun little problem, so though probably the fastest way to answer this exact question is to just write out the heights and add them up, let’s generalize a little.

I’ll define h0 as the initial height from which the ball is dropped, and hi as the height it rises to after the ith bounce. The distance traveled between bounce i and bounce i+1 is thus di=2⋅hi since the ball just goes up and down the same distance. Thus the total distance up to the n-th bounce is:

d=h0+∑n−1i=1di

Next we can use the information that the height after each bounce is a fraction of the previous height. In your specific case that fraction is 34 but in general let’s just call it f . So we then have hi=fhi−1=fih0 and we can rewrite

d=h0+∑n−1i=12fih0=h0[1+2⋅∑n−1i=1fi]=h0[−1+2⋅∑n−1i=0fi]

In the last step here I made the sum start at i = 0 (added 2*1 to the sum so needed to subtract it from the 1 out front), so that we can see this is a standard geometric series. You can figure out how to add up these terms (write out a few terms, multiply by f and write that sum below, then subtract and see nearly everything drop out — or go to ). We get:

d=h0[−1+2⋅1−fn1−f]

In your example, f=34 and n=3 and we get

d=h0[−1+2⋅1−f31−f]=h0[−1+2(1+f+f2)]=h0(1+2f+2f2)=298h0

Explanation:

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