# Launch an 80,000-Pound Sled off an Aircraft Carrier? Sure! You might think this is a video showing the testing of the electromagnetic catapult system for an aircraft carrier, but you are wrong. No, this is a video of a near-perfect example of a real physics homework problem. Yes, that is what it is.

First, you might ask, what is an electromagnetic catapult? So, you have these planes that need to launch from a very short runway. The runway has to be short enough to fit on a boat. The solution to this short runway problem is the catapult. Traditional catapults are essentially a giant steam-powered piston. Steam pushes this piston which pulls the aircraft up to takeoff speed. The electromagnetic catapult is the same thing, except it uses electromagnets.

But I don’t want to talk about the catapult. I want to talk about this sled that is used to test the catapult. Before using this launching device on a real plane, they just put a sled with wheels on the catapult with a mass similar to an actual plane. This sled can’t fly, though, so it gets shot off the edge of the aircraft carrier and crashes into the water. It’s actually pretty cool to watch.

So, why is this a perfect physics problem? Once the sled leaves the carrier deck, there is only one significant force on it—the gravitational force pulling down. This means it is the classic example of projectile motion. Any object that only has the gravitational force acting on it would be a projectile motion problem.

Hold onto your seats. I am going to do this as a full physics problem. Yes, there will be math. Math is your friend. Here is the problem.

An 80,000 pound sled is launched from an aircraft carrier with an initial horizontal speed of 180 knots (92.6 m/s). How far does it travel before hitting the water?

Now that’s a great problem. It’s going to be fun to figure this one out. Of course I got the value for the launch speed from the video. But what about the starting height above the water? Don’t I need that for this problem? Usually, yes. But in this case there is another way to finish the problem.

Let’s talk about projectile motion for a second. The projectile part of this motion starts the instant the sled leaves the carrier and ends right before it hits the water. Remember that both the carrier deck and the water would exert a force on the sled. This means that these extra forces mess up the motion to make it “not projectile motion.”

During the projectile part, however, there are two things that must be true. The velocity of the object in the horizontal direction must be constant. It’s constant because there is no horizontal force to change this horizontal velocity. For the vertical direction, the object will have a constant acceleration (of g = 9.8 meters per second squared) due to the downward gravitational force.

Now for the really cool part about projectile motion. The vertical motion and the horizontal motion of the object are essentially independent. It’s like two separate one-dimensional kinematics problems that only have one thing in common—time. Yes, the time it takes to move in the vertical direction is the same as the time it takes to move in the horizontal direction.

For the vertical motion, the object is accelerating downward such that the position can be determined by the following kinematic equation.

Rhett Allain

In this expression, y is the final vertical position and y0 is the starting vertical position. The coordinate system isn’t real, so you can put the origin wherever you want. However, it might make sense to put the y = 0 meters at the water level. This means the final position is zero, and the initial position is the height of the deck above the water (which I don’t know). Since the sled is launched horizontally, the starting vertical velocity is zero meters per second. That’s useful.

What about the horizontal motion? Since the horizontal acceleration is zero, we get the following kinematic equation.

Rhett Allain

Again, x is the final horizontal position (this is what we want to find) and x0 is the starting position. I’ll just let the starting position be zero meters. Oh, the horizontal velocity will be the 92.6 m/s as stated in the video.

What usually happens next is that I take one of these two equations and solve for the time. I can then use that time in the other equation to solve for what I don’t know. If I had the height of the flight deck, I could use that and the vertical motion to find the time it takes to hit the water. This is the same time in the horizontal direction so that I could solve for the distance it hits the water.

Since I don’t have the flight deck height, I’m going to use another method to get the time. I’m going to get the time from the video. The video of the sled launch isn’t perfect for video analysis since it doesn’t show a nice side view of the motion. But I can still use video analysis to get the time it takes to leave the flight deck and hit the water.

There are many options to get this projectile time. You could use the player in Youtube, but it only counts by seconds (not fractions of a second). I like to use Tracker Video Analysis (it’s free). Actually, this is fairly straightforward for a video analysis problem. I don’t need to find the scale of the video, I just need two times. Assuming the video runs in real time (why wouldn’t it), I get a free fall time of 1.968 seconds.

If I put that time into the horizontal motion equation along with the velocity of 92.6 m/s, I get a horizontal distance of 182.2 meters. That is on the order of two football fields long (a football field is 120 yards long). But there is your answer.

Wait! There’s more! Now that I have the time of free fall, I can use this in the vertical motion to find the height of the flight deck. Plugging in my values, I get a height of 19.98 meters (that’s 65.5 feet).

If that’s not enough physics for you, here are some homework questions.

• Also using video analysis, I found that it took 2.702 seconds for the sled to accelerate while still on the launching system. Using a final speed of 92.6 m/s, what is the acceleration during the launch? How many g’s would this be?
• Assuming the sled starts from rest and the length of the catapult is 105 meters (I measured it), what is the acceleration and final speed? Does this agree with the 92.6 m/s value we used?
• If the sled is 80,000 pounds (as stated in the video), what is the force needed to accelerate this sled?
• How much energy does it take to launch this sled? How many candy bars’ worth of energy is this equivalent to?
• Estimate the power required to launch the sled.
• Suppose you used a 40,000 pound sled instead of the original sled. If the electromagnetic launch system exerts the same force over the same distance, how far will this lighter sled travel before hitting the water? Hint: this is a great question.
• Estimate the air resistance force on the sled during launch. Is it reasonable to assume it is negligible? Hint: I don’t know the answer to this question.