Current time:0:00Total duration:2:39
0 energy points
Studying for a test? Prepare with these 2 lessons on Two-dimensional motion.
See 2 lessons
Optimal angle for a projectile part 3: Horizontal distance as a function of angle (and speed)
Video transcript
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.