A ball is fired from ground level, with a fixed initial velocity, \(u\), at an angle \(\alpha\) to the ground. Assuming no one intercepts the ball, describe how the position at which it lands will vary with the angle of projection.

The sketch below is a reasonable summary of the problem. We are interested in how \(S\) varies with \(\alpha\).

We might like to begin to get a feel for the problem by thinking about some specific cases.

What could you say about the case where \(u=\quantity{5}{m\,s^{-1}}\) and \(\alpha=30^{\circ}, 40^{\circ}, 80^{\circ}\)?

Can you sketch the three cases on the same axes?

- What do you expect to stay the same in all three cases?
- What do you expect to change and in what way?

In order to address the original problem we need to try to use the above experiences to help us to describe the situation in more general terms of \(S\), \(u\) and \(\alpha\).

Thinking carefully about how we calculate the horizontal displacement of the ball for particular values of \(u\) and \(\alpha\), we can apply the same reasoning to determine that \[S=u\cos{\alpha} \times \frac{2u\sin{\alpha}}{g}.\]

Where does each of the factors in this expression come from?

This can be simplified to give \[S=\left(\frac{u^2}{g}\right)\sin{2\alpha}.\]

How did we simplify the expression for \(S\)?

It is often useful to consider the units in an expression like this. We know that the distance, \(S\), should have units m. Does the right-hand side of the formula have the same units?

Representing this equation as a graph might help us to interpret our findings.

- Why is the blue section of the graph highlighted?
- At what angle does the maximum horizontal displacement occur?
- What is the maximum horizontal displacement of the ball?
- How do the specific cases that were suggested above relate to this diagram?

At what angle will the ball land furthest away?

When will the ball land half as far away as this?

How is this pair of graphs connected to each other?

What do they show?

What does this pair of graphs tell you?

Consider the landing position that is furthest away:

- How much faster will the ball need to be fired so that it lands \(\quantity{1}{m}\) further away than this?

(Here we use \(w\) to represent the new initial velocity of the ball)

We know that the landing position of the ball this time will be at \(\frac{w^2}{g}\) and that this must be equal to the original landing position plus \(\quantity{1}{m}\). We can therefore write \[\frac{w^2}{g}=\frac{u^2}{g}+\quantity{1}{m}.\]

We can rearrange this to show the new velocity, \(w\), in terms of the old velocity,\(u\), \[w=\sqrt{u^2+g}.\]

How would we adapt this formula if we wanted the ball to land \(\quantity{2}{m}\) away? What about \(\quantity{d}{m}\) away?

It is often useful to consider the units in a formula like this. We know that velocity, \(w\), should have units \(\quantity{}{m\,s^{-1}}\). Does the right-hand side of the formula have the same units?