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bouncing ball math question?
A ball is dropped from the height of 9ft. The elasticity of the ball is such that it always bounces up one third the distance it has fallen.
(a) Find the total distance the ball has traveled at the instant it hits the ground the fifth time.
(b) Find a formula for the total distance the ball has traveled at the instant it hits the ground the nth time.
I already tried the problem and for (a) i got a mixed fraction 17 and (25/27)ft. the book says the answer is 17 and (8/9) (im off by about 0.1)
for (b) the answer is 18-(1/3)^n-3 but i don't see how it works and how they got it. Some one please explain!
1 Answer
- RaffaeleLv 79 years agoFavorite Answer
(a)
the ball falls 9ft hits the floor and bounces up to 3 ft then goes down for 3ft hits goes up 1 ft
goes down 1 ft...
forgetting first 9ft fall, all others distances are travelled twice by the ball
9 + 2(3 + 1 + 1/3 + 1/9) = 161/9 ft
(b)
9 + 2[3 + 1 + 1/3 + 1/3^2 + --- + 1/3^(n-1)]
9 + 2[3(1 + 1/3 + (1/3)^2 + (1/3)^3 + ... + (1/3)^(n-2))] (*)
remember that a geometric sequence like this
1 + x + x^2 + x^3 + ... + x^(n-1) = (1 - x^n )/(1 - x)
this comes from algebra polynomial factorizing
1 + x = (1 - x^2)/(1 - x)
1 + x + x^2 = (1 - x^3)/(1 - x)
1 + x + x^2 + ... + x^(n-1) = (1 - x^n )/(1 - x)
in our exercise x = 1/3
and
1 + 1/3 + (1/3)^2 + (1/3)^3 + ... + (1/3)^(n-2) = (1 - (1/3)^(n-1) )/(1 - 1/3) =
= (1 - (1/3)^(n-1) )/(2/3) = (3/2) (1 - (1/3)^(n-1) )
so (*) becomes
9 + 2[3 (3/2) (1 - (1/3)^(n-1) ) =
= 9 + 9(1 - (1/3)^(n-1)) = 9(1 + 1 - (1/3)^(n-1) ) =
= 9(2 - (1/3)^(n-1))
to get your book's answer
multiply
18 - 9(1/3)^(n-1)
9 is 3^2 but is also (1/3)^(-2)
18 - (1/3)^(-2)(1/3)^(n-1) =
18 - (1/3)^(n-3)
