# Total Time a ball bounces for from a height of 8 feet and rebounds to a height 5/8

I am trying to solve the following problem.

Let's say a ball is dropped from a height of $8$ feet and rebounds to a height $\frac{5}{8}$ of its previous height at each bounce keeps bouncing forever since it takes infinitely many bounces. This is not true! We examine this idea in this problem.

A. Show that a ball dropped from a height $h$ feet reaches the floor in $\frac{1}{4}\sqrt{h}$ seconds. Then use this result to find the time, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times:

This my my answer for part A. I am not 100% if it is correct.

time at first bounce =$\frac1{4}\sqrt{8}$

time at second bounce =$\frac1{4}\sqrt{8} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)}$

time at third bounce =$\frac1{4}\sqrt{8} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)^2}$

time at fourth bounce = $\frac1{4}\sqrt{8} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)^2} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)^3}$

B. How long, in seconds, has the ball been bouncing when it hits the floor for the nth time (find a closed form expression)? time at nth bounce =

This is my answer for part B. Hopefully it is correct.

$$\frac1{4}\sqrt{8\left(\frac{5}{8}\right)^{n-1}}$$

C. What is the total time that the ball bounces for?

I am unsure how to solve this part of the problem. I would appreciate some help on how to solve it.

• Why is it $\frac{1}{4}*(8)^{0.5}$? – Shahar Apr 7 '14 at 0:07
• that is (1/4) *sqrt(8) – Rohit Tigga Apr 7 '14 at 0:08
• Yeah I know, I'm asking why you're doing that? What in the problem indicates that this is how you get the time? – Shahar Apr 7 '14 at 0:09
• My mistake I edited the problem – Rohit Tigga Apr 7 '14 at 0:14
• Your answer to (A) fails to consider the time upwards. And your answer to (B) is just the time it falls the $n$-th time, not the complete time. – vonbrand Apr 7 '14 at 0:42

A) It firstly asks you to prove that that statement is true. To prove it, we will need a little bit of kinematics. $$\Delta x=v_0t+\frac1{2}at^2$$

No initial velocity, and the only acceleration is gravity ($-9.8 \approx -10 \frac{m}{s^2}$ but we need feet which is: $-32.2 \approx -32\frac{ft}{s^2}$), so we have:

$$-h=\frac1{2}(-32)t^2$$

$$t=\sqrt{\frac1{16}h}=\frac1{4}\sqrt{h}$$

Cool, so now we know that that statement is true. You've got the right pattern, the $h$ will have the following pattern:

$$h=8, 8 \times \frac{5}{8}, \left(8 \times \frac{5}{8}\right) \times \frac{5}{8}, \dots$$

And of course to calculate the total time, you need to add the time it takes for each height. HOWEVER, don't forget about the ball going up as well... So your new times would be:

1st: $\frac1{4}\sqrt{8}$

2nd: $\frac1{4}\sqrt{8} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)}=\frac1{4}\sqrt{8} + \frac1{2}\sqrt{8\left(\frac{5}{8}\right)}$

3rd: $\frac1{4}\sqrt{8} + \frac1{2}\sqrt{8\left(\frac{5}{8}\right)} + \frac1{4}\sqrt{8\left(\frac{5}{8}\right)^2}+\frac1{4}\sqrt{8\left(\frac{5}{8}\right)^2}=\frac1{4}\sqrt{8} + \frac1{2}\sqrt{8\left(\frac{5}{8}\right)} + \frac1{2}\sqrt{8\left(\frac{5}{8}\right)^2}$

4th: $\frac1{4}\sqrt{8} + \frac1{2}\sqrt{8\left(\frac{5}{8}\right)} + \frac1{2}\sqrt{8\left(\frac{5}{8}\right)^2}+\frac1{4}\sqrt{8\left(\frac{5}{8}\right)^3}+\frac1{4}\sqrt{8\left(\frac{5}{8}\right)^3}=\frac1{4}\sqrt{8} + \frac1{2}\sqrt{8\left(\frac{5}{8}\right)} + \frac1{2}\sqrt{8\left(\frac{5}{8}\right)^2} + \frac1{2}\sqrt{8\left(\frac{5}{8}\right)^3}$

b) You kind of got it right. It's asking how long it has been bouncing at $n$th bounce. What you got is the time for that individual bounce. However, you need the total time. Hence, you need a series. I'd model it like so:

$$\text{total time (at bounce }n)=\left(\sum_{i=1}^{n}\frac1{2}\sqrt{8\left(\frac{5}{8}\right)^{i-1}}\right) - \frac1{4}\sqrt{8}$$

c) This is a convergent geometric series. The sigma term converges to:

$$-\frac{4-\sqrt{2}}{\sqrt{10}-4}$$

When combining with the right term, you get:

$$-\frac1{\sqrt{2}}-\frac{4\sqrt{2}}{\sqrt{10}-4}\approx 6.0456 \; s$$

That is the answer for the total amount of time.

• Need to acount for the time spent travelling upwards (same as downwards for the next bounce). – vonbrand Apr 7 '14 at 0:43
• Your final sum is geometric... – vonbrand Apr 7 '14 at 0:44
• @vonbrand That's right. I just noticed that. – Shahar Apr 7 '14 at 0:45

Let's take from (B) The first fall is for 8 feet, then the ball climbs $8 \cdot \frac{5}{8}$ and falls $8 \cdot \frac{5}{8}$, and so on. The time the first fall is $4 \sqrt{8}$, the second fall (and the climb before it) take $2 \cdot 4 \sqrt{8 \frac{5}{8}}$, and so on. The time to get to the $n$-th time to the floor is (add in a $0$-th climb for regularity of the sum, substract it later): \begin{align} \sum_{0 \le k \le n} 2 \cdot 4 \sqrt{8 \left( \frac{5}{8} \right)^k} - 4 \sqrt{8} &= 16 \sqrt{2} \cdot \sum_{0 \le k \le n} \sqrt{\left( \frac{5}{8} \right)^k} - 8 \sqrt{2} \\ &= 16 \sqrt{2} \cdot \frac{1 - \left( \sqrt{5/8} \right)^n}{1 - \sqrt{5/8}} - 8 \sqrt{2} \\ &= 128 \cdot \frac{1 - \left( \sqrt{5/8} \right)^n}{2\sqrt{2} - \sqrt{5}} - 8 \sqrt{2} \end{align}

After infinite bounces the power vanishes: $$\frac{128}{2 \sqrt{2} - \sqrt{5}} = \frac{128 (2 \sqrt{2} + \sqrt{5})}{13}$$

[MathJax complains about the above align, but I don't see any problems. My network is very flaky right now, so I can't rule out a network hickup.]