A ball is thrown horizontally at a speed of \(\quantity{2}{m\,s^{-1}}\) from a point \(\quantity{1.25}{m}\) above \(O\) which is on a flat horizontal surface. After each bounce it reaches a height which is \(75\%\) of its maximum height after the previous bounce. Ignore air resistance and take the acceleration of gravity, \(g\), to be \(\quantity{10}{m\,s^{-2}}\).

  • Sketch the trajectory of the ball.

In between bounces, the ball is a projectile in free-fall so it’s trajectory will follow a parabola. After each bounce it joins a different parabola, with a maximum at \(75\%\) of the height of the previous one.

A sequence of parts of parabolas

This diagram might raise a number of questions which probably can’t be answered without doing some calculations that we’ll get onto later.

  • Are all the parabolic sections the same “width”?
  • If not, how are they related?
  • How does the time taken for each section vary?
  • Does this sequence of parabolas continue indefinitely?
  • Calculate \(t_n\), the time the ball is in the air between the \(n^\text{th}\) and \((n+1)^\text{th}\) bounces.

Between bounces, the ball is a projectile moving under gravity. We can treat the horizontal and vertical motions as independent of one another. The horizontal motion has no effect on the time between bounces.

Vertically, let’s start by thinking about the first section, before the first bounce. The ball starts with zero vertical velocity and we want to know the time taken to fall under gravity through a distance \(\quantity{\frac{5}{4}}{m}\). \[\begin{align*} s&=ut+\frac{1}{2}at^2 \\ \frac{5}{4} &=0+\frac{1}{2}10t^2 \\ t^2&=\frac{1}{4} \end{align*}\]

So the time for this section, \(t_0=\frac{1}{2}\).

We can do a similar calculation for \(t_n\) (for \(n\ge1\)), using the reduced maximum height. The height after the \(n^\text{th}\) bounce will be \[\frac{5}{4}\times\left(\frac{3}{4}\right)^n.\] So the constant acceleration formula for the downward journey gives us \[\frac{5}{4}\times\left(\frac{3}{4}\right)^n = 0+\frac{1}{2}10t^2\] and so the time for the whole section (up and down) will be \[t_n=2\times\sqrt{\frac{1}{5}\times\frac{5}{4}\times\left(\frac{3}{4}\right)^n} = \left(\frac{\sqrt{3}}{2}\right)^n.\]

Why did we include a factor of \(2\) at the start of this expression?

This starts to answer some of our earlier questions. Each section of the trajectory takes less time than the previous one, by a factor of \(\sqrt{3}/2\). Assuming the horizontal velocity is a constant \(\quantity{2}{m\,s^{-1}}\), what is the distance between successive bounces?

  • How far from \(O\) does the ball get before it stops bouncing?

This might seem like a strange question – does it not keep bouncing forever?

We could take a pragmatic approach such as, “in reality a ball won’t appear to be bouncing when the height gets below a few centimetres. Find out how many bounces that will take and add up the times or distances for that many bounces.” Instead, we have considered a theoretical ball for which any bounce can be measured…

The times for the sections of the trajectory form a sequence, \[\frac{1}{2}, \frac{\sqrt{3}}{2}, \left(\frac{\sqrt{3}}{2}\right)^2, \left(\frac{\sqrt{3}}{2}\right)^3, \cdots\] Apart from the first term, \(t_0\), this is a geometric sequence with first term, \(t_1=\frac{\sqrt{3}}{2}\) and common ratio \(\frac{\sqrt{3}}{2}\).

As the common ratio is less than one we can calculate the sum to infinity of its terms. \[\sum_{n=0}^\infty t_n = \frac{1}{2} + \frac{\sqrt{3}/2}{1-\sqrt{3}/2} = \frac{7}{2}+2\sqrt{3} \approx7.0\]

Surprisingly, this tells us that although the ball bounces an infinite number of times, the time between bounces gets shorter and shorter, so that after \(7\) seconds all the bounces have been completed and the ball no longer moves vertically at all!

Assuming the horizontal velocity is a constant \(\quantity{2}{m\,s^{-1}}\), the ball will be about \(\quantity{14}{m}\) from \(O\) when it stops bouncing. After that it will be rolling along the ground.

If we had based our calculation on a finite number of bounces we would have found a shorter distance than this.

In reality, the horizontal velocity would probably not be constant, but how it changes at each bounce is hard to model.

trajectory of the ball showing decreasing loops converging to a point 14 meters from O

Some more questions to think about…

  • What assumptions were made in the model of this situation? How might the answer be different if we had made different assumptions?
  • Can you think of any other effects that might change the outcome?
  • If we tried this experimentally what do you think would actually happen? Do you think this model would be accurate?
  • If we varied the initial height, how would the total bouncing time change?
  • If we drew a curve through the maximum points of all the parabolic sections, what shape would the curve be? [Express \(x\) and \(y\) in terms of \(n\)…]