# What is the function of the top point of a bouncing ball?

A ball is thrown away as parallel to x axis from M(0,h) point with speed V . After each jumping on x axis , it can reach half of previous height as shown in the figure.(Assume that no any air friction and the ball is very small.)

Question-$1$:

Find f(x) that passes from of top points ?

Question-$2$:

I have just a claim about the system but I need to prove that if we throw away the ball from a point of $f(x)$ as parallel to x axis with the speed of $V_{initial}=V\sqrt{\frac{f(x)}{h}}$, the maximum points of the ball will be on same $f(x)$. How can the claim be proved or disproved?

My attempt to find f(x)

$f(0)=h$

$f(x_1+x_2)=h/2$

$f(x_1+2x_2+x_3)=h/4$

$f(x_1+2x_2+2x_3+x_4)=h/8$ .

.

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$x_1=vt$

$mgh=\frac{1}{2}mV^2$

$h=\frac{1}{2}g t^2$

After first bouncing , On first top point: N $(x_1+x_2,h/2)$:

$mgh/2=\frac{1}{2}mV_1^2$

$V_1^2=V^2/2$

$V_1=\frac{V}{\sqrt{2}}$

$h/2=\frac{1}{2}g t_2^2$

$t_2^2=t^2/2$

$t_2=\frac{t}{\sqrt{2}}$

$f(x_1+x_2)=h/2$

$f(Vt+t_2\frac{V}{\sqrt{2}})=h/2$

$f(Vt+\frac{t}{\sqrt{2}}\frac{V}{\sqrt{2}})=h/2$

$f(Vt(1+1/2))=h/2$

After second bouncing, On second top point: O $(x1+2x2+x_3,h/4)$:

$mgh/4=\frac{1}{2}mV_2^2$

$V_2^2=V^2/4$

$V_2=\frac{V}{2}$

$h/4=\frac{1}{2}g t_3^2$

$t_3^2=t^2/4$

$t_3=\frac{t}{2}$

$f(x_1+2x_2+x_3)=h/4$

$f(Vt+Vt+V_3t_3)=h/4$

$f(Vt+Vt+Vt/4)=h/4$

$f(Vt(1+1+1/4))=h/4$

On 3th top point:

$f(Vt(1+1+1/2+1/8))=h/8$

On 4th top point:

$f(Vt(1+1+1/2+1/4+1/16))=h/16$ .

.

On nth top point: $n--->\infty$

$f(3Vt)=0$

How can I find $y=f(x)$

Thanks for answers

EDIT: I have found $f(x)$. MrBrody is right. The function is a line.

my Proof: On nth top point:

$f(Vt.a(n))=h/2^n$

$a(0)=1$

$a(1)=1+1/2=1+1-1/2$

$a(2)=1+1+1/4=1+1+1/2-1/4$

$a(3)=1+1+1/2+1/8=1+1+1/2+1/4-1/8$

$a(4)=1+1+1/2+1/4+1/16=1+1+1/2+1/4+1/8-1/16$ .

.

$a(n)=1+1+1/2+1/4+....1/2^{n-1}-1/2^{n}$

$a(n)=1+\frac{1-(1/2)^{n})}{1-1/2}-1/2^{n}=1+2(1-(1/2)^{n})-1/2^{n}=3-\frac{3}{2^n}$

On nth top point:

$$f(Vt[3-\frac{3}{2^n}])=\frac{h}{2^n}$$

where $t=\sqrt{\frac{2h}{g}}$

$Vt[3-\frac{3}{2^n}]=x$

$3-\frac{3}{2^n}=\frac{x}{Vt}$

$1-\frac{1}{2^n}=\frac{x}{3Vt}$

$1-\frac{x}{3Vt}=\frac{1}{2^n}$

$$f(x)=\frac{h}{2^n}=h(1-\frac{x}{3Vt})=h(1-\frac{x}{3V\sqrt{\frac{2h}{g}}})$$

Now I am trying to prove or disprove my claim in second question. If I find the solution I will post it.

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Nice problem. Is this from a book, or did the professor make it up? – ja72 May 23 '12 at 23:19
no any book just a my idea – Mathlover May 23 '12 at 23:24
+1 for the idea. – ja72 May 24 '12 at 3:03

## 1 Answer

The key parameter here seems to be how are distributed the losses at each bounce (from what you said, there is no air friction here): how much translational and vertical speed are lost in the transition before/after the bounce. From your graphic, it seems like the hypothesis is that there is "reflection" of the ball: same angle when arriving/leaving the ground, only the norm of the speed decreases, the ratio |V_y|/|V_x| being conserved. From that, you can prove that each parabola is similar to the previous one, same ratio height/length on x axis for each one. From this you indeed get a straight line, whith its slope determined by the 2 first points, (0,h) and (x1+x1/2, h/2). Then you only need to determine x1 to determine the slope of f(x), and hence f(x) completely. I hope I didn't miss anything, as this hypothesis on the repartition of the losses on V_x and V_y at each bounce is crucial to solve the problem!

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