# What Goes Up …

I was a mediocre student in maths before I started at a university. But in university, I discovered the amazing things that maths can do. Maths can be used to describe all sorts of physical processes and this can be used to control those processes. Things like the cruise control in your car or the flight controls of an aircraft or spacecraft need a mathematical model of the thing being controlled. One of the first examples used to introduce students to modelling (that is mathematically describing something), is the simple act of throwing a ball into the air.

Now to start modelling this process from scratch requires calculus which I haven’t covered yet. But I will give you the final result and we will work with that.

If a tennis ball is hit straight up in the air, there are two main forces acting on it after it leaves the racquet: gravity and drag from the air. Though the air drag can be modelled as well, it complicates the model so it is usually assumed to be negligible when introducing this to students. The effect of gravity is to reduce the initial speed given to the ball by 9.8 m/s every second. So if the ball has an initial speed of 60 m/s, after 1 second its speed is 60 – 9.8 = 50.2 m/s. After the next second, its speed is 50.2 – 9.8 = 40.4 m/s, and so on. By the way, I am using metric units here. The same thing can be done with American units where the effect of gravity reduces the speed of the ball by 32 ft/s every second. But as most of the world uses metric, we’ll stay with that.

So let’s stick with the initial velocity of the ball as 60 m/s when it leaves the racket. Now the first thing to do is agree on a coordinate system. It’s natural to agree that up is positive and down is negative. We’ll also agree that time starts at 0 as soon as the ball leaves the racket and that the height in meters at the point where the ball leaves the racquet is also 0. We’ll call this the ground level. So using calculus with the force of gravity and the initial speed of the ball (ignoring air drag), we can get three equations that describe the motion of the ball: one equation for its acceleration, one for its velocity, and one for its height from the ground. With a = acceleration, v = velocity, h = height, and  t = time in seconds, these equations are:

$\begin{array}{l} {{a}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}}\\ {{v}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}}\\ {{h}\hspace{0.33em}{=}\hspace{0.33em}{-}{4}{.}{9}{t}^{2}\hspace{0.33em}{+}\hspace{0.33em}{60}{t}} \end{array}$

With these equations, we can answer the following questions:

1. When will the ball hit the ground?
2. How fast is the ball travelling when it is at its maximum height?
3. How high does the ball go?
4. How fast is the ball going when it hits the ground?

Before I answer these questions, let’s look at the graph of the height equation. Now before, we talked about x and y values on a graph. But for physical processes, we can use different letters that are more meaningful. Instead of x, we will use t for time, and instead of y, we will use h for height. The graph of this equation is below:

The curve is an upside down parabola. The graph goes on forever below the t-axis, but I’m only show the part of the graph that makes physical sense.

The graph and the equation make sense at t = 0 seconds as h is 0 on the graph at t = 0, and if you let t = 0 in the equation, you also get h = 0.

So you can see that the ball goes up, reaches a maximum height, then falls back to the ground. To answer question 1, from the graph, it looks like the ball hits the ground a little over 12 seconds because that is where the graph shows h = 0 again. To find the exact value, we must set the height equation equal to zero, and find the times when h = 0. This will require us to factor the equation and use the null factor law (explained in a previous post):

$\begin{array}{l} {{h}\hspace{0.33em}{=}\hspace{0.33em}{-}{4}{.}{9}{t}^{2}\hspace{0.33em}{+}\hspace{0.33em}{60}{t}\hspace{0.33em}{=}\hspace{0.33em}{t}{(}{-}{4}{.}{9}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}}\\ {\Longrightarrow\hspace{0.33em}{-}{4}{.}{9}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{-}{4}{.}{9}{t}\hspace{0.33em}{=}\hspace{0.33em}{-}{60}}\\ {\Longrightarrow{t}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{60}}{{-}{4}{.}{9}}\hspace{0.33em}{=}\hspace{0.33em}{12}{.}{245}\hspace{0.33em}{\mathrm{seconds}}} \end{array}$

So the ball travels for 12.245 seconds before it hits the ground (that’s a powerful tennis player!). Notice that there is also another solution, t = 0, which is when the ball is initially hit.

Question 2 can be answered by the physics of the problem and this answer will help us answer question 3. As explained, the ball has an initial velocity of 60 m/s, but is continually slowing down due to gravity. When it reaches its maximum height, the ball reverses direction then goes back down. At the maximum height, the velocity is 0 because velocity is positive going up and negative going down, so it must be 0 right when the ball reverses direction.

Question 3 can be answered using the answer to question 2. If we set the velocity equation to zero, that will give us the time that the ball is at maximum height. We then use this time in the height equation to find the height at that time:

$\begin{array}{l} {{v}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}{t}\hspace{0.33em}{+}\hspace{0.33em}{60}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\Longrightarrow\hspace{0.33em}{-}{9}{.}{8}{t}\hspace{0.33em}{=}\hspace{0.33em}{-}{60}}\\ {\Longrightarrow\hspace{0.33em}{t}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{60}}{{-}{9}{.}{8}}\hspace{0.33em}{=}\hspace{0.33em}{6}{.}{122}\hspace{0.33em}{\mathrm{seconds}}}\\ {{h}\hspace{0.33em}{=}\hspace{0.33em}{-}{4}{.}{9}{(}{6}{.}{122}{)}^{2}\hspace{0.33em}{+}\hspace{0.33em}{60}{(}{6}{.}{122}{)}\hspace{0.33em}{=}\hspace{0.33em}{183}{.}{67}\hspace{0.33em}{\mathrm{meters}}} \end{array}$

We also could have solved this by noticing that a parabola is symmetric and the maximum height would occur halfway between 0 and 12.245 seconds. This also would have given us t = 6.122 seconds to use in the height equation.

The last question is answered by using the time that the ball hits the ground in the velocity equation:

${v}\hspace{0.33em}{=}\hspace{0.33em}{-}{9}{.}{8}{(}{12}{.}{245}{)}\hspace{0.33em}{+}\hspace{0.33em}{60}\hspace{0.33em}{=}\hspace{0.33em}{-}{60}\hspace{0.33em}{\mathrm{m}}{/}{\mathrm{s}}$

Notice that the velocity is the same as when the ball was first hit except it is negative since it is now going down instead of up.

Isn’t it amazing how we can find out all sorts of things about throwing a ball without actually doing it!