Question

A ball is dropped from a height of 10 feet and bounces. Each bounce is 34 of the height of the bounce before. Thus, after the ball hits the floor for the first time, the ball rises to a height of 10(34)=7.5 feet, and after it hits the floor for the second time, it rises to a height of 7.5(34)=10(34)2=5.625 feet. (Assume that there is no air resistance.)

Answer 1

Let a be original height

r the ratio of bounce

so we get

a

after first bounce ar

after second bounce = ar^2

a) after n bounce = ar^n

b) distance travelled first bounce = a+ar

distance travelled second bounce = a+ ar+ ar+ ar^2

( ar going up and down so twice)

distance travelled third bounce = a+ar+ar+ar^2+ar^2+ar^3

distance travelled fourth bounce = a+ar+ar+ar^2+ar^2+ar^3+ar^3+ar^4

c) a+ 2(ar+ar^2+ar^3+...ar^(n-1) ) + ar^n

= 2(a+ ar+ar^2+ar^3+...ar^(n-1) + ar^n ) - a - ar^n

= 2a(1 -r^(n+1) )/(1-r) - a - ar^n

substitute a = 10 and r = 3/4 to get answers