# Projectile Motion: Example Part 1

by Jared Rovny

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00:01 We have here a projectile motion example problem that will show us how to solve a complicated question and it asks, if you launched a cannon ball at a hundred kilometers, or sorry, a hundred meters per second at an angle 45 degrees with the horizontal, how far will this cannon ball go and what maximum height will the cannon ball reach? What you should do is first is use the equations of motion and the principles I just described about how these projectile motion problems work and give it a shot.

00:29 See if you can do this, try to figure out how far it would go by using your horizontal equations or how high its maximum height would be by using the vertical equations and sometimes needing to use these equations together.

00:40 Once you've given that a shot here's one way we can go about solving this problem.

00:46 So first what I'm going to do is draw a quick schematic of what this looks like.

00:52 We have cannon ball and we're gonna launch it at a velocity, v.

00:56 And we're given in this problem an angle theta which is 45 degrees in our case.

01:01 As we discussed, supposing they're launch and then this gonna land somewhere, and this asks what is the distance, the total distance that this cannon ball is going to go which I'm calling d. As we mentioned the first thing we need to do is break this vector into its two components - a horizontal and a vertical component.

01:22 So first this horizontal component as we saw, is v times the cosine of theta and the vertical component, will be v times the sine of theta. With these two we're ready to solve the horizontal and vertical components and I'm going to keep these separate. So I'm going to write the horizontal here.

01:42 And I'll try to keep this in separate columns and I will write the vertical over here.

01:47 So first let's look at the horizontal because if we're looking at this problem and we see the distance variable d over here and we wanna find how far it goes, this indicates to us that we're talking about the horizontal direction because we just care about how far this object goes.

02:03 What we know from our equations of motion, that the positions equals whatever position that it was originally in, plus the initial velocity in the x direction, in this case times the time plus 1/2 the acceleration in the x direction times the time squared where importantly we've said that there's no horizontal acceleration.

02:24 So really all we have to say that is x minus x zero by rearranging this equation is equal to the initial velocity in the x direction times the time so this would be our answer.

02:36 We would be pretty well done with this problem.

02:37 We just have x minus x zero is equal to the velocity times the time so all we have to do is multiply velocity by however long the object is in the air and we're done.

02:46 We know how far the object went.

02:47 As you might guess get we have one problem here which is that the problem does not tell us anything about the time. We have no idea how long this cannon ball is flying through the air.

02:55 So we're going to need to resort to the vertical component of this problem to find how long the ball is in the air before we can go back to the horizontal component of this problem and solve by plugging in the time. So let's do that now.

03:07 In the vertical direction we can say a number of things. So for example, let's start cataloging.

03:14 We know that the initial position in the vertical direction is zero because we're gonna take a snapshot with our ball starting here, and ending here.

03:21 So the vertical position of this ball is zero initially and the vertical position of the ball finally is also zero because it's also in the ground so it has no vertical height.

03:31 So we know the initial and final position of the ball.

03:35 We also know that the velocity of the ball in the y direction is the velocity times the sine of theta and therefore we can plug in for some of our equations of motions so for example our velocity equation would say that the final velocity of our cannon ball is equal to the initial velocity plus the acceleration times the time.

03:57 And again having used our equations of motion we know that this is initial velocity minus g times the time.

04:05 But this equation since we don't technically know what the final velocity is, we could use a symmetry argument and say that the initial velocity upwards is going to be the same but opposite of the final velocity downwards if we're going to nor anything that could take away from our velocity but technically we don't know this, and it might be tricky on an exam to try to remember such a symmetry argument.

04:24 So let's try to not use this first equation, though again you could as I've just described, let's instead use our full position equation where your final position equals you initial position plus your velocity in the y direction times time plus 1/2 the acceleration times the time squared, and let's see if we can instead use this equation.

04:44 As you can see cataloging our knowns and unknowns, we know the final position is zero.

04:49 We know the initial position is zero. We know what the velocity is and we know the initial velocity and we know what the acceleration is, meaning the only variable in this equation we don't know is time which is perfect for us because we like to use the time to solve for the horizontal part.

05:04 So let's go ahead and see what this equation gives us.

05:06 We know that the final position is zero, we know the initial position is zero so we can rearrange this equation and say that 1/2 a t squared is equal to minus having subtracted it from both sides the initial velocity in the y direction times time.

05:22 Dividing both sides by time, we have 1/2 a times t equals minus initial velocity in the y direction, and we can solve for the time by multiplying both side by 2 and dividing by a, so we have minus 2 initial velocity in the y direction, divided by the acceleration which is again, minus g, cancelling the minus signs, we have 2 times the velocity in the y direction over g. So this is our time.

05:47 This is great, because we know all of these variables and we solved for the time which means we can plug this back into our horizontal equation. Doing that we're almost done.

05:56 We found that the distance x minus x zero, is equal to the initial velocity in the x direction times this time variable which we've just solved for using the vertical equations, 2v zero y over g. So we put 2v zero y over g and now we can plug in our variables and we are finished with this problem just by plugging this in.

06:19 So then we have 2 velocity in the x direction, velocity in the y direction divided by g and these are 2 and we've solved these velocity in the x direction initially is the cosine of theta and the velocity in the y direction is the sine of theta, so we have 2v square, and we have a sine of theta and we have a cosine of theta and these are all divided by g and then all we have to do to finish this problem is plug in these numbers, keeping this horizontal and vertical separate we have two.

06:52 We were given a velocity of 100 kilometers per hour or meters per second rather squared and then we have sine of theta and cosine of theta.

07:00 Now we're gonna have to do a little bit of trigonometry here and we will review these angles and I recommend you do as well when we show the slide at least, which is at the sine of 45, is equal to the square root of 2 over 2.

07:10 And it turns out this is the same thing as the cosine of 45, at 45 degrees sine and cosine are the same.

07:15 And then also we have value of g. As I've said, g is equal to 9.8 meters per second squared.

07:24 To one significant figure, what I'm going to do for most of these lectures is actually assume a value g that is 10. There's two reasons for doing this.

07:32 One is that just as I'm doing this I won't be spending a lot of time going through very, my calculations that are relevant to the point of this problem.

07:40 But the second reason and the more important reason is that when you're going through problems like this on any exam you're taking related to physics you'll want to be able to quickly find approximations, because this will help you quickly diagnose problems and quickly figure out which answer is close and if you ever need the extra accuracy, you can always go back in and plug in the more accurate value of 9.8 if you'd like.

08:01 So we're just gonna use g being approximately 10 to one significant figure here.

08:06 So plugging these values in and recognizing that square root of 2 over 2 squared, is just 2 over 4 or 1/2, then we finish this problem by plugging in the sine of theta square root of 2 over 2, cosine of theta is also square root of two over two. So multiplying them together, and we square the value and got in 1/2 and then we divide it by this value of ten.

08:32 So the 2 here, we'll cancel this 1/2 here. We have a hundred squared which gives us four zeros divided by 1 zero gives us 3 zeros and so this is a quick way to think about things again using scientific notation, and we get a final answer of 1000 meters as the range of our object.

The lecture Projectile Motion: Example Part 1 by Jared Rovny is from the course Translational Motion.

### Included Quiz Questions

1. 2V²sinθcosθ/g
2. 2Vsinθcosθ/g
3. 2V²sinθ/g
4. 2V²sinθcosθ
5. V²sinθcosθ/g
1. θ = 45 degrees
2. θ = 90 degrees
3. θ = 180 degrees
4. θ = 60 degrees
5. θ = 70 degrees
1. (V/g)sinθ
2. (2V/g)sinθ
3. V/g
4. (2V/g)sinθcosθ
5. (2V²/g)sinθcosθ
1. y = y₀ + Vsinθ t - gt²/2
2. y = y₀ + Vsinθ t
3. y = y₀ + Vsinθ t - gt²
4. y = y₀ + Vt - gt²/2
5. y = y₀ + Vsinθ t + gt²/2

### Author of lecture Projectile Motion: Example Part 1 ### Customer reviews

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Very nice
By Neuer N. on 12. October 2021 for Projectile Motion: Example Part 1

Very good explanation. I thank Lecturio for making this true.