# How High Can You Jump? From Randall Munroe’s “*How To”*

I’m currently reading Randall Munroe’s book, *How To *(Penguin Random House). Personally, I love anything from Randall Munroe — you might know him from his awesome xkcd comics.

The basic idea of the book is to take some crazy (or even normal) questions and explain how to do them. For instance, this question was “how to jump high”. Of course, it’s just a book — and with that, he has to skip over some of the details. But that’s why I’m here. I’m going to fill in all the missing details of some of these equations.

Here is the one I want to start with. The idea is that if you are moving with some speed, you can redirect that velocity upward and send yourself into the air. This would be the case of running on a skateboard and then hitting a vertical ramp. Munroe then gives the following equation (which I have reproduced with LaTeX).

Yes, this was a challenge to reproduce. I used Apple Keynote software (which does LaTeX a little different) and this is the code for that expression.

`\mathrm{height}=\frac{\mathrm{speed}^2}{2 \times \mathrm{acceleration}\medspace \mathrm{of}\medspace \mathrm{gravity}}=\frac{\left(\frac{100 \medspace \mathrm{meters}}{10 \medspace \mathrm{seconds}}\right)}{2 \times 9.805 \medspace \frac{\mathrm{m}}{\mathrm{s}^2}}=5.10 \medspace \mathrm{meters}`

You don’t need that code, I just wanted to show you that I can reproduce the equation.

Before I show you where this equation comes from, I have some comments:

- There’s a typo. When he puts in the numbers for the velocity, he dropped the “squared” part.
- This equation is in metric units — YAY!
- I’m not a big fan of “acceleration of gravity”. I think a better option would be to either just use “g” or “the gravitational field”. Actually, I normally see “acceleration DUE TO gravity” — which I still don’t really like.
- I think it’s OK to use speed instead of velocity — it can get tricky though.

Let’s begin with the derivation.

Suppose you have a ball and you toss it straight up with an initial velocity. After it leaves your hand, there is only one significant force acting on it — the downward gravitational force. Near the surface of the Earth, the gravitational force pulls straight down and has a magnitude that depends on both the mass of the object (m) and the gravitational field (g).

Yes, both the force and the gravitational field are vectors. Since m*g has to be in units of Newtons, the magnitude of the gravitational field would be:

A couple of comments:

- Since g is the magnitude of the gravitational field, it’s positive. Don’t write g = -9.8 N/kg or you will look like a physics N00B.
- A Newton is a kg*m/s². So, if you divide a Newton by a kilogram, you get m/s² — which is the unit for acceleration.

The net force on the ball is m*g and according to Newton’s second law (in just the y-direction):

So, the ball will have an acceleration of -9.8 m/s². If it starts with a positive velocity(v0) with that acceleration of -g, how high will it go? Of course you could just use the magic words of “kinematic equations” — and that’s cool and all. However, I want to start from something more basic: the definition of acceleration (in 1D) and average velocity. Note: I’m going to drop the y-subscript for the velocity and acceleration — just for simplicity.

Let’s start with acceleration.

Since the acceleration is the rate of change of velocity, we can write it as the change in velocity (final velocity v2 minus initial velocity v1) divided by the time interval. From the gravitational force, this acceleration is -g. The negative just means that the change in velocity is in the negative direction. If the ball is moving up (positive y-velocity), it will slow down. If the ball is moving down (negative velocity) it will speed up.

But does this acceleration equation give us what we need? No, we want the maximum height of the ball, this doesn’t have any positions in the equation. It’s still useful though. The initial velocity is v1 and the final velocity is zero. Yes, when the ball gets to its highest point, it has to transition from moving up (positive velocity) to down (negative velocity) — that means it has to cross through zero velocity.

With that, I can solve for the time it takes to get to this highest point.

Caution: this expression ONLY works for the case where the final velocity is zero. But notice that it is indeed a positive time interval. That’s good.

Now let’s go to the definition of the average velocity (in one dimension).

See, there are two definitions of the average velocity. It’s the rate of change of position — but it’s also the average velocity (initial plus final divided by 2). This assumes that the velocity is changing at a constant rate (constant acceleration).

I want to find the change in position (Δy) — but this is just the total height of the ball (or person). Also, v2 is still zero at the highest point. Solving for the change in position using the two average velocity definitions:

Next I can just substitute in my expression for the time interval and I get:

Boom. That’s the same thing as from the book.

But what about the numerical value for the height? Don’t worry, I will make a python calculator for you. Here is the code, and this is what it looks like.

That’s it. That’s the equation from *How To*.