Asked by PDF
A rubber ball rebounds 3/5 of the height from which it falls. If it is initially dropped from a height of 30 ft. What total vertical distance does it travel before coming to rest?
Answers
Answered by
Reiny
Make a diagram
distance during first bounce - 15 ft
distance during 2nd bounce - 2(15)(3/5)
distance during 3rd bounce = 2(15)(3/5)^2
distance during 4th bounce = 2(15)(3/5)^3
etc
from the 2nd term on, we have a geometric series
where a = 2(15)(3/5) = 18 and r = 3/5
and the want Sum∞
sum∞ = 15 + a/(1-r)
= 15 + 18/(1-3/5)
= 15 + 18/(2/5)
= 15 + 45 = 60
distance during first bounce - 15 ft
distance during 2nd bounce - 2(15)(3/5)
distance during 3rd bounce = 2(15)(3/5)^2
distance during 4th bounce = 2(15)(3/5)^3
etc
from the 2nd term on, we have a geometric series
where a = 2(15)(3/5) = 18 and r = 3/5
and the want Sum∞
sum∞ = 15 + a/(1-r)
= 15 + 18/(1-3/5)
= 15 + 18/(2/5)
= 15 + 45 = 60
Answered by
Abel
a sub 1 = 30
r = 3/5
S = (r) / (1-r)
S = {(30ft) / [(1)-(3/5)]}(3/5)
S = 150ft / 5-3 = 150ft / 2 = 70ft
r = 3/5
S = (r) / (1-r)
S = {(30ft) / [(1)-(3/5)]}(3/5)
S = 150ft / 5-3 = 150ft / 2 = 70ft
Answered by
Michelle
60
Answered by
Myla
120 ft
Answered by
Anonymous
120...
30+ summation(X to infinity)(30*2*(3/5)^x)
30+90=120
30+ summation(X to infinity)(30*2*(3/5)^x)
30+90=120
Answered by
anonymous
S = 30 + 2 [15/(1-3/5)]
S = 30 + 2(15/2/5)
S = 30 +75
S = 105 ft
S = 30 + 2(15/2/5)
S = 30 +75
S = 105 ft
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