How Fast Was this Crazy Car Jump?

I should start a series titled: things you should never do but that I will analyze. Here is my latest entry. Oh, my alternate title for this post is "YEEEEE HAW!"

This is uber dangerous and illegal. You shouldn't try anything like this. Just to be clear, it is bad. Now on to the physics.

Question: how fast was this car going?

Let me start with a simplified diagram.

If I assume that the air resistance is small (which probably isn't a great assumption), I can treat this car like a projectile motion problem. This would be a projectile motion problem with a twist, a twist that makes it interesting. If the car is launched at that angle, how far down the angled road will it land (I will call this distance s). Actually, I will estimate the launch angle (θ), the road landing angle (α) and the landing distance. From these, I will calculate the launch speed.

Projectile Motion

If you want a detailed review of projectile motion, check this out. Here is the short version:

For an object that is moving only due to the gravitational force, it will only have an acceleration in the y-direction. This means that in the x- and y-directions, I can write:

Just to be clear, I am calling the start of the launch time t = 0 seconds. Now, normally, this would be a fairly easy problem to solve. You would use the y-equation to solve for the time and then use that same time in the x-direction. However, the problem here is that the final y-position is not zero. It depends on how far horizontally it moves.

Let me go ahead and say that the car starts at x = 0 m, and y = 0 m (so the origin is at the launch point). In this case, I can write an expression for the equation of the landing road.

This is just the equation of a straight line that passes through the origin. The slope is negative of the tangent of the slope angle. I used prime notation so that the x' and x values wouldn't be confused.

Now, back to the vertical projectile motion equation. Instead of saying that I will solve for the time when the car gets to y = 0 meters, I will say the final y is the value of the 'road equation'. (remember that y₀ is zero now that I set the origin at the launch location)

I also know an expression for the x-direction. It will take the same time as the y-direction, so I can write:

Now, I can combine those two equations (by substituting for x') to get an expression with just time in it:

With a little bit of algebra, I can get this:

With this time, I can get the x-coordinate of the landing position.

However, I want the distance (s) down the road that the car would land. If I know x, and I know the angle α then s would be:

Now I can put the expression for x back in and solve for v₀

Ok, this looks a little complicated. Let me do some usual checks to make sure I didn't make a mistake.

• Does it have the correct units? Check.
• What if it is a flat road (α = 0)? In this case it should reduce to the plain old projectile motion on a flat surface. Check.
• What if the car shoots straight up (θ = π/2)? It should land at x = 0. Check.

Just because these things check out, doesn't mean it is correct. However, I feel more comfortable now.

Estimation of car speed

Take a look at this shot.

From that, I am totally going to estimate a road angle of α = 10 degrees and a launch angle of θ = 5 degrees. After watching the videos like 8 times in a row, I am going to estimate s of about 6 car lengths. If the car is about 5 meters long, then s would be about 30 meters. Sure, I could have done some video analysis on this, but I wanted to try something different.

Using the above numbers, I get a car speed of about 39.7 m/s or 88.8 mph. 88 miles per hour!