# Ask Uncle Colin: Jumping

Dear Uncle Colin,

I want to jump over the Grand Canyon on a motorbike. How fast would I need to go?

Extreme Velocity, Extreme Landing

Hi, EVEL, and thanks for your message!

According to Wikipedia, the Grand Canyon varies in width from four miles to 18 miles. I’d hesitate to recommend you go through with this plan, but if you do, I’d suggest doing it where the width is closer to four miles.

I’m going to assume a couple of things:

- You have one of those amazing new suits that drops your air resistance to effectively zero, so that I can neglect it
- You’re going to take off at an angle of 45 degrees to the horizontal and land at the same height.
- The distance you want to jump is 7,000m.

Suppose you take off at a speed of $u\sqrt{2}$ m/s (the $\sqrt{2}$ makes the sums easier later). Taking off at 45 degrees, you’ll be travelling with both a horizontal and vertical velocity of $u$ m/s.

At time $t$ after take-off, your displacement from your take-off point will be $ut$ m horizontally, and $ut - \frac{1}{2}gt^2$ m vertically.

We want the first of those to be 7,000, and the second to be 0, so we need to solve:

- $ut = 7000$
- $\frac{1}{2} t(2u - gt) = 0$

Since you’ll be in the air for longer than 0 seconds, it’s clear that the solution we want for the second equation is $2u=gt$, or $t = \frac{2u}{g}$.

Substituting into the first equation gives $2u^2 = 7000g$.

Let’s take $g$ to be 10m/s^2, so we have $u^2 = 35,000$, which makes $u \approx 190$m/s and your takeoff speed something like 270m/s.

That works out to be a little shy of 600mph, which – even for a Montana rider such as yourself – strikes me as unreasonable and/or imprudent. The current world speed record for a motorbike is slightly under 400mph.

I mean, best of luck and all, I hope you stick the landing – and I hope that helps!

- Uncle Colin