A ball is dropped from a height of ten feet and bounces... math problem?

Question

I have been slaving at this problem for days. I am just looking for some help.

...Each bounce is 3/4 of the height of the previous bounce. Assume acceleration due to gravity is 32ft/sec^2. Recall the position and velocity functions.
a) Find an expression for the height the ball achieves after the nth bounce.

b) Find an expression for the total vertical distance the ball has traveled after the first, second, third, fourth, and fifth times.

c) Find an expression for the total vertical distance the ball has traveled after the nth bounce.

d) Show that a ball dropped from a height of h feet reaches the ground in 1/4(h)^(1/2)seconds.

e) Show that the ball in part a stops bouncing after (1/4) (√10)+(1/2) (√10) (√3/4) (1/(1-(√3/4))) seconds

Answer 1

After the first bounce, the ball will be 3/4 of 10 ft = 7.5 feet.



So the sequence is 7.5, 3/4(7.5), ...



It is a geometric sequence with first term a = 7.5 and ratio 3/4:



So the height after the nth bounce is h = ar^(n-1) = 7.5(3/4)^(n-1)

You can also express it as

h(n) = 10(3/4)^n



b)

First, lets start at the point th ball hits the ground for the first time.

From there on, it will travel once up some height, and then once down that same height.

And then it will bounce up 3/4 the previous heigh, and come down that same distance.



After the first bounce, it goes up 7.5, and comes down 7.5, then goes up 3/4 * 7.5 and down 3/4 * 7.5

We want the sum of the distances travelled in this sequence, which starts at 15.



For the sum of a geometric sequence, the formula is:

S(n) = a(1 - r^n )/(1 - r)

However, this formula by itself will give the distance travelled n bounces after the first bounce. So the correct formula, so the adjusted formula is:

S(n) = a(1 - r^(n - 1) )/(1 - r)

Subbing in the first term and ratio:

S(n) = 15( 1 - (3/4)^(n - 1) )/(1 - 3/4)

Add 10 ft to this to count the distance travelled before the ball bounced for the first time:



S(n) = 10 + 15( 1 - (3/4)^(n-1) ) / ( 1 - 3/4)



Now find S(6) for the answer.



c) The formula above answers this



d) The function for the height of the ball as a function of time (up till the first bounce) is:

y(t) = h - 1/2 g t²

g is 16.



When the ball reaches the ground, the height y is 0, so:

0 = h - 1/2 (32) t²

0 = h - 16t²

16t² = h

t² = h/16

t = √h / 4



e) Don't feel like thinking anymore. :p

Answer 2

Bounce #1 --- (9/10)10 Bounce #2 --- (9/10)(9/10)10 = (9/10)^210 Bounce #3 --- (9/10)(9/10)(9/10)10 = (9/10)^310 Bounce #4 --- (9/10)(9/10)(9/10)(9/10)10 = (9/10)^410 . . . Bounce #n --- (9/10)^n10 Taking Lim (9/10)^n10 as n --> ∞ The limits evaluates to zero. So, it doesn't diverge. Now take the SUM: => ∑(9/10)^n10 as n --> ∞ The sum evaluates to 90 [units]. It's preposterous that the ball would bounce infinitely. What planet do people live on anyway? hth

Answer 3

HEIGHT = 10 FT.

g = 32 ft/sec^2

Distance travelled before first bounce = 10 ft.

Distance travelled before 2nd bounce = 2*(3/4)*10 ft.

Distance travelled before 3rd bounce = 2*[(3/4)^2]*10 ft.

Distance travelled before 4th bounce = 2*[(3/4)^3]*10 ft.

Distance travelled before 5th bounce = 2*[(3/4)^4]*10 ft.

.....

.....

Distance travelled before nth bounce = 2*[(3/4)^(n-1)]*10 ft.

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