Here is a great question from a reader (rephrased a bit):

I have in my possession a device from Noah Basketball and it measures the approach angle of a basketball at the rim using a video camera.

The manufacturers claim to have studied well over ten thousand players at various levels. They claim that a medium high arc of 43 to 47 degrees (depending on the shooter’ height) will result in an optimal shot. What I’m trying to figure out is the relationship between the approach angle at the rim and:

varying the release angle and launch speed of the shotthe height of the playerthe distance from the basket

Let’s get to work.

## Assumptions

I think that in this situation, it is safe to assume that air resistance is negligible. Boy, that would be a pain in the rear if I had to consider air resistance. So, whether it is true or not, I am not going to add air resistance.

One other thing. I am not going to look at the side to side variation in a shot. I will just assume the shooter can aim straight. If you are the coach and your players are shooting straight, maybe you could practice shooting straight.

I am not sure if I am going to consider backboard shots.

## Projectile Motion Primer

Let me try something a little different here. I usually post all the equation details. Perhaps many people just skip over those steps. For now, let me just say that for projectile motion, we have the following two equations for the motion in the x- and y-directions:

Here *x* and *y* are clearly changing with time. Also, I took one shortcut. I used *t*. This assumes that at *t* = 0 seconds, the object is at position *x*_{0} and *y*_{0}.

The general solution for projectile motion is to put in the stuff you know. Then use one of the above equations to solve for the time. This time can then be used in the other equation.

Ok, now some variable to use in this basketball situation. Let me start with this diagram:

Actually, I just realized something. If I put the origin at the starting location of the ball, then I can get rid of one of the heights. Let me call the difference in height of the starting and ending points *h*. If I want to refer to the starting height of the ball, I will call that *p* (for person).

This means that my two kinematic equations become:

Now what? Well, I could solve for a number of things – but really I am looking for relationships between the variables. Honestly, it is pretty simple to solve for the initial speed needed to hit a certain spot if you know everything else. It is not so trivial to solve for the angle needed if you know the speed. To make things a little easier, I will switch to numerical mode. And to do that, I will need some starting values.

- The height of the rim is 3.05 meters above the ground. Let me assume a release height of 2 meters. This means that
*h*would be 1.05 meters. - What about the distance from the basket? The three point line is around 7 meters (depending on the type of court). How about I start with a distance of 5.5 meters.
- What ranges of initial speeds seem reasonable? I will start with something low like 5 m/s and move up to about 15 m/s. I doubt I will need to go much faster than that.
- The basketball rim has a diameter of about 45 cm. The basketball has a radius of about 12 cm.

Here is the plan: use standard projectile motion calculations to model where the ball will go given an initial velocity and launch angle. Next, see if this trajectory will take it through a basketball goal. Pretty simple, right? Well, the idea is simple but the calculation can take a while.

If I vary the launch angles from 35° to 70° and I vary the launch speed from 7 m/s to 11 m/s, which combinations would result in a goal? Remember that I am not looking at backboard shots or ones that spin around the rim. These are just plain old through the hoop shots. Here is what I get:

Which (by the way) agrees with the data I posted in this previous post on basketball shots.

But what does this plot show us? First, it shows that I was a fool for including speeds lower than 7.6 m/s. Next, it looks like a launch angle of about 50 degrees is pretty nice. Why? First, this angle corresponds to the lowest launch speed. Second, it seems like this is the thickest part of the curve. So, if you vary your launch speed some, you will still make the shot.

But does this answer the original question? I think not. Let me make a plot of starting angle vs. entering angle for all of these shots.

This shows that there seems to be a fairly linear correlation between the initial throwing angle and the angle the ball has when it hits the goal (for this same distance and height from the goal). So, perhaps this is one of the answers to the questions. If the best launch angle is around 50 degrees, this would correspond to one “entering angle” of about -40°. The video basketball computer can’t really see the launch angle, but it can see the final angle.

## Final Note:

Do you know what is really cool? Even though I can look at projectile motion and calculate optimal launch angles and stuff, I can’t really shoot any better than the average person. On the other hand, an NBA pro could take a shot from a whole bunch of different locations and make many of them. Some of these NBA players have no clue about projectile motion (although surely some do).

So, how do people make these types of shots? If you say “muscle memory” or something like that, I don’t like it. It could be muscle memory if they always shot from the same place with the same initial velocity and angle. But these players shoot all over the place. They jump and shoot. They move the side then shoot. Crazy.

Authors: