Asked by heyo ❄
17. A rubber ball dropped on a hard surface takes a sequence of bounces, each one 3 / 5 as high as the preceding one. If this ball is dropped from a height of 10 feet, what is the total vertical distance it has traveled at the time when it hits the surface for its fifth bounce?
23 7/125 feet
36 14/125 feet
43 111/125 feet
46 14/125 feet
23 7/125 feet
36 14/125 feet
43 111/125 feet
46 14/125 feet
Answers
Answered by
mathhelper
Make a sketch showing up to the fifth bounce and your
total distance = 10 + 2(10)(3/5) + 2(10)(3/5)^2 + 2(10)(3/5)^3 + 2(10)(3/5)^4
notice the first term does not fit the geometric series, so let's fix it
= [ 20(3/5) + 2(10)(3/5) + 2(10)(3/5)^2 + 2(10)(3/5)^3 + 2(10)(3/5)^4 ] - 10
a = 20
r = 3/5
n = 5
total distance = sum(5) - 10
= 20( 1 - (3/5)^5 )/(1-3/5)
= 20(5/2)(1 - 243/3125)
= 50(2882/3125)
= 5764/125 , I see that in your choices
total distance = 10 + 2(10)(3/5) + 2(10)(3/5)^2 + 2(10)(3/5)^3 + 2(10)(3/5)^4
notice the first term does not fit the geometric series, so let's fix it
= [ 20(3/5) + 2(10)(3/5) + 2(10)(3/5)^2 + 2(10)(3/5)^3 + 2(10)(3/5)^4 ] - 10
a = 20
r = 3/5
n = 5
total distance = sum(5) - 10
= 20( 1 - (3/5)^5 )/(1-3/5)
= 20(5/2)(1 - 243/3125)
= 50(2882/3125)
= 5764/125 , I see that in your choices
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