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Question 1092224: A tennis ball is dropped from a height of 40m and bounces in place. Each time
it strikes the ground, it bounces to 40% of its previous height. Find the total
vertical distance traveled by the ball if it is stopped immediately on the 10th
time it hits the ground.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The distance is 40 + 2*16 (1-r^8)/(1-r)+ 40*r^10=40+32(0.9993/0.6)+0.004=93.302 feet. ANSWER
The first distance is kept out of the series. Once the ball rises, there are 8 complete cycles with a ratio of 2/5 or 0.8, and 1 more time that it hits the ground for the 10th time. The 2 is the factor that doubles the distance and the 16 is the height the ball is after the first bounce.
32(1-r^8/(1-r)), and 1-r^8=0.99934
Check
1st drops 40 comes up 16=56
2nd drops 16 comes up 6.4=22.4 (78.4)
third drops 6.4 comes up 2.56=8.96 (87.36)
fourth drops 2.56 comes up 1.024 (90.94)
fifth drops 1.02 comes up .41=1.43 (92.37)
sixth drops .41 comes up .16=0.57 (92.94))
seventh drops .16 and comes up .064=0.22 (93.16)
eighth drops .064 and comes up .026=0.09 (93.255)
ninth drop .026 and comes up about 0.01=0.03 (93.29)
tenth drops 0.01 (93.30).
The formula is easier if the question asked for the vertical distance to return to the hand after the 10th drop.
While it is tempting to round 1-0.4^n to 1, it adds error.
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