Optimal angle for a projectile
Optimal angle for a projectile part 3: Horizontal distance as a function of angle (and speed)
We now know how long the object is going to be in the air, so we're ready to figure out how far it's going to travel. So we can just go back to kind of the core formula in all of really kinematics, all of kind of projectile motion or mechanical physical problems, and that's distance is equal to rate times time. Now, we're talking about the horizontal distance. So our distance is going to be equal to-- what's our rate in the horizontal direction? We care about horizontal distance traveled, so our rate needs to be the horizontal component of the velocity, or the magnitude of the horizontal component of the velocity. And we figured that out in the first video. That is s cosine of theta. So let's write that down right here. So our rate is s cosine of theta. And how long will we be traveling at this horizontal speed? Well, we'll be going at that speed as long as we are in the air. So how long are we in the air? Well, we figured that out in the last video. We're going to be in the air this long-- 2 s sine of theta divided by g. So the time is going to be 2 s sine of theta over g. So the total distance we're going to travel, pretty straightforward, rate times time. It's just the product of these two things. And we could put all of the constants out front, so it's a little bit clearer that it's a function of theta. So we can write that the distance traveled-- let me do that same green. The distance traveled as a function of theta is equal to-- I'll do that in this blue. This s times 2s divided by g is-- I'll do it in a neutral color actually. This s times 2s divided by g is 2 times s squared over g. So 2s squared over g times cosine of theta times sine of theta. So now we have a general function. You give me an angle that I'm going to shoot something off at and you give me the magnitude of its velocity, and you give me the acceleration of gravity. I guess if we were on some other planet, who knows? And I will tell you exactly what the horizontal distance is.