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lilavasa [31]
3 years ago
8

A ball is dropped from a height of 32 feet. it bounces and rebounds 80% of the height from which it was falling

Physics
2 answers:
n200080 [17]3 years ago
7 0

Answer:

d = 32*( 1 - 0.8^n) / 0.2 = 160*( 1 - 0.8^n)

Explanation:

Given:

- The initial height of the ball h_o = 32 ft

- The successive decrease in height after every bounce = 0.8*h

Find:

- The expression relating the total distance traveled by the ball for nth number of bounce.

Solution:

- The distance traveled by the ball upto n = 1, is = 32 ft

- The distance traveled by the ball upto n = 2, is = 32 + 2*32*.8 = 83.2

- The distance traveled by the ball upto n = 3, is = 83.2 + 2*32*.8*.8 = 124.16

- We can look for a pattern for the total distance traveled by geometric progression is as such:

                            d = a*( 1 - r^n) / ( 1 - r )

Where, r = 0.8 , and a = 32

                            d = 32*( 1 - 0.8^n) / 0.2 = 160*( 1 - 0.8^n)

Where, n = 1 , 2 , 3 , 4 , ......

Step2247 [10]3 years ago
5 0

Answer:

If it falls from 32 feet, how could the distance be 29 feet? Twelve bounces later, 58 is "obviously" incorrect as well. Eliminate those two before you do anything else.

The total distance up until that the nth bounce is  

Sn = (32 - 32(.8)12) / (1 - .8) = 149.004883722... = 149

Explanation:

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