General mathematical solution to this problem? Code solution given

Question

I'd like to count the number of times a ball will bounce above a certain height. Imagine a ball dropped from a building, and which bounces by a factor of 0,66. How many times it will reach the 2nd floor window at 1.5 meters high from the floor?

I solved it using code, but I thought that maybe there is a general solution to this. Unfortunately my maths courses are far now and I can't quite remember the techniques.

Assumptions are:

• height > window's height
• 1 > bounce rate > 0

Here is the code I used to solve it:

def number_of_bounces(height, bounce_rate, window_height):
    """
    Calculates the number of times the ball will pass in front of the window
    """
    if height < window_height:
        return 0
    next_bounce = height * bounce_rate
    if next_bounce < window_height:
        return 1

    passage = 1
    while next_bounce > window_height:
        passage += 2 # Once upward, once downward
        next_bounce = next_bounce * bounce_rate

    return passage

Any idea for a general solution to this problem?

Edit : Tests for this problem : Initial height : 2 / Bounce rate : 0.5 / Windows height : 1 / Expected result : 1

Initial height : 3 / Bounce rate : 0.66 / Windows height : 1.5 / Expected result : 3

Initial height : 30 / Bounce rate : 0.66 / Windows height : 1.5 / Expected result : 1

Initial height : 30 / Bounce rate : 0.75 / Windows height : 1.5 / Expected result : 21

Answer 1

We have by energy

$$ \cases{ mgh_0 = \frac 12 m v_0^2\\ mgh_1 = \frac 12 m v_1^2\\ \vdots\\ mgh_k = \frac 12 m v_k^2 } $$

and also $v_{k+1} = \lambda v_k$ then

$$ h_{k+1} = \lambda^2 h_k,\ \ \ h_0 = H $$

and solving this recurrence

$$ h_k = H\lambda^{2k} $$

but $h_k = \frac g2 t_k^2$ or $t_k = \sqrt{\frac 2g H}\lambda^k$ and finally

$$ T_k = t_0+2\sum_{j=1}^{j=k} t_j $$

$$ T_k = \sqrt{\frac 2g H}+2\sqrt{\frac 2g H}\sum_{j=1}^k\lambda^j = \sqrt{\frac 2g H}+2\lambda\sqrt{\frac 2g H}\left(\frac{1-\lambda^k}{1-\lambda}\right) $$

NOTE

The "number of seen bounces" can be obtained as follows:

From $h_k = H\lambda^{2k}\ge \bar{h}$ we conclude that

$$ k = \text{floor}\left[\frac 12\log_{\lambda}\left(\frac{\bar{h}}{H}\right)\right]+1 $$

here $\lambda\lt 1$

or

$$ k = \text{floor}\left[\frac 12\ln\left(\frac{\bar{h}}{H}\right)/\ln(\lambda)\right]+1 $$

Attached a MATHEMATICA script showing the bouncing process.

parms = {H -> 3, h -> 1.5, mu -> 0.66, g -> 9.81};
parms = {H -> 30, h -> 1.5, mu -> 0.66, g -> 9.81};
parms = {H -> 30, h -> 1.5, mu -> 0.75, g -> 9.81};
tmax = 15;
ODE = {y''[t] == -g, y'[0] == 0, y[0] == H, WhenEvent[y[t] == 0, y'[t] -> -mu y'[t]]} /. parms
soly = NDSolve[ODE, y, {t, 0, tmax}][[1]];
Plot[{y[t] /. soly, (h /. parms)}, {t, 0, tmax}, PlotRange -> {0, H /. parms}]

Answer 2

Let $H$ be the initial height and $h$ the height of the window.

You are looking for the quantity $2n$ such that

$$H 0.66^{n+1}<h\le H 0.66^{n}$$

Taking logarithms, you will have

$$\ln(H)+(n+1) \ln(0.66) < \ln(h) \le \ln(H)+n \ln(0.66)$$

Now form $\frac{\ln(h)-\ln(H)}{\ln(0.66)}$. What can you conclude ?

Answer 3

The ball is dropped with null velocity at $t=0, z=h$.

The ball dynamics is given by $$ \left \lbrace \begin{array}{ccl} v_z &=& - gt \\ z &=& -\frac12 gt^2 +h \end{array} \right. $$ Simple computations show that the velocity at impact time $t_1=\sqrt{2h/g}$ is $ v_1^-=-\sqrt{2gh} $.

The velocity just after bounce is $v_1^+ = -\eta v_1^-$.

We know that for an object sent from the ground at positive vertical velocity $v$ the maximum height being reached is $h_\mathrm{max}={v^2}/{2g}$.

Call $v_n$ the velocity at $n$th bounce It holds $v_n^+ = \eta^n \sqrt{2gh}$ which yields the maximum height of $\eta^{2n}\cdot h$

Let $h=30$ $\eta=0.75$, we have $0.75^{2*5}\cdot 30 = 1.68$ and $0.75^{2*6}\cdot 30 = 0.95$ This gives 10+1=11 crossings as shown in the Mathematica plot above. So the answer must be 11 and not 21 ?