# Projectiles: What’s the optimal angle at which to throw something? (to maximize distance) Wow, a real life formula that uses the double angle trig. identities!  As you can see, the distance a projectile will travel is a function of:  velocity, gravity, and the launch angle.

First, a quick fraction review:  First, recall that $$\frac{1}{1000}$$ is a lot smaller than $$\frac{1}{10}$$.  Conversely, we can also agree that $$\frac{1}{100}$$ is a lot smaller than $$\frac{99}{100}$$.  ie: The bigger the denominator (and/or smaller the numerator), the lower the value of the (positive) fraction.

So, since v is in the numerator, the distance traveled (d) increases directly with velocity (in a big way, since it’s squared)  Next, since g is in the denominator, the distance traveled decreases as gravity increases.  (Makes sense, right?) This is a graph of all Sin(x) values from 0 to 360.  The x-axis is divided into quadrants (0, 90, 180, 270, 360). Notice in the graph that Sin(x) rises from 0 to 1 as x rises from 0 to 90 degrees.  Then, it drops from 1 back to 0 as x rises from 90 to 180 degrees.

Refer back to the double angle $$Sin(2\theta)$$ in the original formula up top.  So, as x rises from 0 to 45 degrees, 2x actually rises from 0 to 90, and the $$Sin(2\theta)$$ value is increasing.  But, as x continues to rise from 45 to 90 degrees, 2x rises from 90 to 180, which means the $$Sin(2\theta)$$ value is now decreasing.

So, what’s the ideal angle to throw something?  The one that maximizes the value of $$Sin(2\theta)$$, since it’s a multiplier in the projectile formula.  Well, as you can see in the graph, Sin(90) = 1, the highest possible value for Sin(x).  So, the ideal launch degree is x = 45 (which puts 2x at 90).

So, now you know why the ideal angle in these video games is 45 degrees, and have an inkling of how programmers create classic games like these: ## One thought on “Projectiles: What’s the optimal angle at which to throw something? (to maximize distance)”

1. Tony on said:

The example of the old Olympic video game with the long jumper is ironic, because I was just playing that precise game with my daughter over the weekend! You have to jump 6 meters in order to make “qualify” in the event and we were barely making it. I’ll have to share the result with her so we can compete with Carl Lewis next time. I’ll let you know if we set a new personal record. I used to compete in the long jump in H.S., but this fact was never explained to me. I should have asked my Physics teacher for his help instead of the rah rah track coach! I was too short for high jump, but the optimal launch angle must be different and perhaps an off-shoot of this equation.