A ball is dropped from a height of 17 feet and bounces. Suppose that each bounce

Question

A ball is dropped from a height of 17 feet and bounces. Suppose that each bounce is 5/8 of the height of the bounce before. Thus, after the ball hits the floor for the first time, it rises to a height of 17(58)=10.62517(58)=10.625 feet, etc. (Assume g=32ft/s^2 g=32ft/s^2 and no air resistance.)

A. Find an expression for the height, in feet, to which the ball rises after it hits the floor for the nth time:
hn=

B. Find an expression for the total vertical distance the ball has traveled, in feet, when it hits the floor for the first, second, third and fourth times:
first time: D =  
second time: D =  
third time: D =  
fourth time: D =

C. Find an expression, in closed form, for the total vertical distance the ball has traveled when it hits the floor for the nnth time.
Dn=

Explanation / Answer

Let a be original height

r the ratio of bounce

so we get

after first bounce = a*r

after second bounce = a*r^2

(a) after n bounce = a*r^n

(b) distance travelled first bounce = a+a*r

distance travelled second bounce = a + a*r+ a*r+ a*r^2

distance travelled third bounce = a+a*r+a*r+a*r^2+a*r^2+a*r^3

distance travelled fourth bounce = a+a*r+a*r+a*r^2+a*r^2+a*r^3+a*r^3+a*r^4

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

Dn = 2(a+ a*r+a*r^2+a*r^3+...a*r^(n-1) + a*r^n ) - a - a*r^n

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

substitute a = 10 and r = 5/8 to get answers

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