I told you I was going to be back.

So last time I left you with a incomplete analysis of a jump shot that Stephen Curry took back in January.  Well I am back to it. If you don't remember the shot here it is.


For a more complete analysis here it is.....



So last time we looked at the ball's trajectory and we analysed how fast Curry had to throw the ball in order to get it to go in the hoop.  We didn't however talk about the physics behind the throw. While the ball is in the air it is what we call projectile.  Now the only force that is acting on a projectile is gravity. (If we ignore air resistance) It turns out that, at sea level you can calculate this force very easily using Newton's universal law of gravity.

Where m₁ is the mass of the earth in kilograms, m₂ is the mass of the object in kilograms, G is the gravitational constant 6.674×10⁻¹¹ N⋅m²/kg²  and r is the distance between center of the two masses.

Now we can use another one of Newton's laws F=ma to find the acceleration due to gravity near the earth's surface. It turns out that if we do the calculation, the acceleration is 9.8 m/s^2.  The cool thing about this is that it doesn't mater what that object is, the acceleration is always the same.  This makes our job of modeling those projectiles much easier.  Now we could go through the mathematics of two dimensional projectiles here but I think that I have covered enough physics for one blog post.  I do want do show you the model of Curry's shot so I wrote a little bit of code to show you how projectiles can be modeled in a computer.  So here it is... 

If you hit the play button at the top of the code you can see my model of Curry's shot.  I do want you to look at the code first so that you can see how I modeled it.

Thanks

Tobin Giraud