A ball is dropped from a height of 19 feet and bounces. Suppose that each bounce

Question

A ball is dropped from a height of 19 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 19 ({5/8}) = 11.875 feet, etc. (Assume g = 32 hbox{ft/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: h_n =

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 nth time.

D_n =

Explanation / Answer

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 = 19 and r = 5/8 to get answers

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