Optimal angle for a projectile
Optimal angle for a projectile part 1
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- Let's say we're going to shoot some object into
- the air at an angle.
- Let's say its speed is s and the angle at which we shoot
- it, the angle above the horizontal is theta.
- What I want to do in this video is figure out how far
- this object is going to travel as a function of the angle and
- as a function of the speed, but we're going to assume that
- we're given the speed.
- That that's a bit of a constant.
- So if this is the ground right here, we want to figure out
- how far this thing is going to travel.
- So you can imagine, it's going to travel in this parabolic
- path and land at some point out there.
- And so if this is at distance 0, we could call this distance
- out here distance d.
- Now whenever you do any problem like this where you're
- shooting something off at an angle, the best first step is
- to break down that vector.
- Remember, a vector is something that has magnitude
- and direction.
- The magnitude is s.
- Maybe feet per second or miles per hour.
- And the direction is theta.
- So if you have s and theta, you're giving me a vector.
- And so what you want to do is you want to break this vector
- down into its vertical and horizontal components first
- and then deal with them separately.
- One, to help you figure out how long you're in the air.
- And then, the other to figure out how far
- you actually travel.
- So let me make a big version of the vector right there.
- Once again, the magnitude of the vector is s.
- So you could imagine that the length of this arrow is s.
- And this angle right here is theta.
- And to break it down into its horizontal and vertical
- components, we just set up a right triangle and just use
- our basic trig ratios.
- So let me do that.
- So this is the ground right there.
- I can drop a vertical from the tip of that arrow to set up a
- right triangle.
- And the length of the-- or the magnitude of the vertical
- component of our velocity is going to be this
- length right here.
- That is going to be-- you could imagine, the length of
- that is going to be our vertical speed.
- So this is our vertical speed.
- Maybe I'll just call that the speed sub vertical.
- And then, this right here, the length of this part of the
- triangle-- let me do that in a different color.
- The length of this part of the triangle is going to be our
- horizontal speed, or the component of this velocity in
- the horizontal direction.
- And I use this word velocity when I specify
- a speed and a direction.
- Speed is just the magnitude of the velocity.
- So the magnitude of this side is going to be speed
- horizontal.
- And to figure it out, you literally use
- our basic trig ratios.
- So we have a right triangle.
- This is the hypotenuse.
- And we could write down soh cah toa up here.
- Let me write it down in yellow.
- soh cah toa.
- And this tells us that sine is opposite over hypotenuse,
- cosine is adjacent over hypotenuse and tangent is
- opposite over adjacent.
- So let's see what we can do.
- We're assuming we know theta, we know s.
- We want to figure out what the vertical and the horizontal
- components are.
- So what's the vertical component going to be?
- Well the vertical component is opposite this theta.
- But we know the hypotenuse is s, so we could use sine
- because that deals with the opposite and the hypotenuse.
- And the sine function tells us that sine of theta-- actually,
- let me do this in green since we're doing all the vertical
- stuff in green.
- Sine of theta is going to be equal to opposite, which is
- the magnitude of our vertical velocity.
- So the opposite side is this side right here, over our
- hypotenuse.
- And our hypotenuse is the speed s.
- And so if we want to solve for our vertical velocity or the
- vertical component of our velocity, we multiply both
- sides of this equation by s.
- So you get s sine of theta is equal to the vertical
- component of our velocity, s sine of theta.
- And now for the horizontal component we do the same
- thing, but we don't use sine anymore.
- This is now adjacent to the angle.
- So cosine deals with the adjacent side and the
- hypotenuse.
- So we could say that the cosine of theta is equal to
- the adjacent side to the angle, that is the horizontal
- speed, over the hypotenuse.
- The hypotenuse is this length right here, over s.
- So if we want to solve for the horizontal speed or the
- horizontal component or the magnitude of the horizontal
- component, we'd just multiply both sides times s.
- And you get s cosine of theta is equal to
- the horizontal component.
- So we now know how fast we are travelling in this direction,
- in the horizontal component.
- We know that that is going to be s cosine of theta.
- And we know in the vertical direction-- let me do that in
- the vertical direction, the magnitude is s sine of theta.
- It is s sine of theta.
- So now that we've broken up into the two components, we're
- ready to figure out how long we're going to be in the air.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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