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Optimal angle for a projectile part 1: Components of initial velocity

You want a projectile to fly as far as possible, at which angle should you launch it? We'll start with formulas for the initial velocity. Created by Sal Khan.

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  • blobby green style avatar for user Dorien Baudewyns
    what's the diffrence between speed and velocity?!
    (14 votes)
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    • female robot grace style avatar for user M
      also, if you are talking about average speed and average velocity, average speed can never be zero, whereas average velocity can be zero. This is because Average Speed = total distance traveled/total time taken and average velocity = displacement/time taken. Displacement can be zero because it is the shortest distance between two point and in case the moving object returns to it's initial position then displacement is zero. Hope that helps and adds to your question.
      (13 votes)
  • blobby green style avatar for user Olivia Gibson
    What do you do if the problem given does not have an angle? What equation do you use to find the angle?
    (8 votes)
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    • female robot grace style avatar for user Niki
      I think you can use reverse trigonometry, if it gives you two sides you can use Pythagorean because it'll be a right triangle, and you can plug in the side lengths for reverse trigonometry.
      (4 votes)
  • hopper cool style avatar for user ★✮✶Arsh Pervez✶✮★
    What do scos and ssin mean?
    (9 votes)
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  • piceratops tree style avatar for user Noelle Kline
    i noticed that you use the theta symbol and i would like to know what it means.
    (3 votes)
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  • duskpin ultimate style avatar for user UnrealDreamer989
    How would the problem change if you are given an initial or final height of something other than 0?
    (6 votes)
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  • blobby green style avatar for user aarshsingh11
    how do you derive the maximum height reached by the object?
    (3 votes)
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    • orange juice squid orange style avatar for user rudymik
      I'm assuming you want the answer, and not a mathematical proof.
      For just a quick review, the three most important equations in projectile motion are:
      Δx = vt + (1/2)at^2
      vfinal = vinitial+ at
      (vfinal)^2 = (vinitial)^2 + 2aΔx

      The maximum height of a projectile can be found from the formula (v)^2 = 2aΔy, where v is the initial vertical velocity of the projectile and a is the acceleration (most often 9.8 m/s/s, or "little g"). The formula can be rearranged to find the vertical displacement (maximum height):
      Δy = v/2a
      (2 votes)
  • mr pants pink style avatar for user Aldo Elias
    how do you write maximum height?
    (2 votes)
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  • aqualine ultimate style avatar for user knott.lm
    Would the distance be any different if there was any wind, or would it change if the power of what was being launched was changed? And just a quick thing as well, the said "projectile" was not specified if it was or wasn't round, cause if it was round wouldn't it roll making the distance even farther than it was when it first hit the ground? Unless it was say a box that was launched, I do belive I have just confuesed myself in keeping this straight, I don't quite understand stuff like this to well, so answers would be nice. :D
    (2 votes)
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  • piceratops ultimate style avatar for user Nathan
    at Sal says that the direction is theta, but the diagram depicts theta as being the angle at which the projectile is thrown, Which is the right one?
    (1 vote)
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    • male robot hal style avatar for user Henry Meehan
      They're the same thing! If an angle relative to the horizontal axis (x-axis) is given along with a line (in this case vector s), then we know the direction of that line. If you think about it, these two bits of information (angle and magnitude) can tell us a lot about an object's trajectory. Theta is just a term for an angle for which we don't have a precise figure.
      (2 votes)
  • starky ultimate style avatar for user Toska Pi
    Hi everyone! I was wondering - the shape of the trajectory is parabolic, so why do we make calculations using a triangle with straight edges (because the trajectory isn't a straight line). This question especially applies to where the object is launched directly from the ground. Do we just imagine a triangle a bit behind where the trajectory starts?
    (1 vote)
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Video transcript

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.