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?

Answer 1

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

Answer 2

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

Answer 3

60

Answer 4

120 ft

Answer 5

120...
30+ summation(X to infinity)(30*2*(3/5)^x)
30+90=120

Answer 6

S = 30 + 2 [15/(1-3/5)]
S = 30 + 2(15/2/5)
S = 30 +75
S = 105 ft