# Homework Help: Projectile Motion- How to Calculate Displacement

1. Jun 6, 2014

### Jai

1. The problem statement, all variables and given/known data

I am wanting to calculate the Displacement of a projectile when it is launched at various angles, however all I have is the Initial velocity and the acceleration due to gravity.

3. The attempt at a solution

first attempted to approach this by calculating the horizontal and Vertical initial velocity. I then realised that since air resistance is being neglected the horizontal acceleration should be 0. That is as far as I got and I'm not even sure if I was on the right track.

Sorry if it is obvious but it is just not coming to me...
-Thanks

Last edited: Jun 6, 2014
2. Jun 6, 2014

### Nathanael

Welcome to physics forums.

Have you tried figuring out how long it will be in the air?

Do you know how to go about figuring that out?

3. Jun 6, 2014

### Jai

Thanks for the reply. Yes I did consider that and try using the equations t=s/v, V=u+at, S=ut+1/2at^2 and V^2=u^2+2as but none of these seemed to work. They are the only equations I have been taught involving time.

4. Jun 6, 2014

### Nathanael

Do you know what the vertical velocity will be when it hits the ground? (Assuming you know the initial vertical velocity)

If you know the initial vertical velocity and the final vertical velocity, then you know what the total change in vertical velocity is.

And you also know the vertical acceleration, so you could potentially figure out the time that way.
(time = change in velocity / acceleration)

5. Jun 6, 2014

### Jai

Possibly. I did work something out but I am not sure if it is correct. the initial vertical velocity is 4.038 m/s^2. I then subtracted the acceleration due to gravity giving me -5.77m/s^2. Sorry if this is all wrong I've just really confused myself here...

6. Jun 6, 2014

### Nathanael

Those are units of acceleration, so that can't be right.

Are we assuming that the projectile is launched from ground level?

If so:
Will the time it takes to go up (and reach maximum height) be less than, greater than, or equal to, the time that it takes to fall back down to the ground?

7. Jun 6, 2014

### Jai

Yes, it is launched from ground level. Won't the differences in time depend on the angle?

8. Jun 6, 2014

### Nathanael

Well the angle only effects how much of the initial velocity is "converted" into vertical velocity and horiztonal velocity. ("Converted" is not the right word. But do you understand what I'm trying to say?)

When we look at the time it takes to reach the maximum height, (and the time it takes to fall from that maximum height) we are only interested in the vertical velocity. We can basically pretend we're throwing something straight up in the air.

If I throw something in the air will the time it takes to rise be less than, greater than, or equal to, the time that it takes to fall?
(To fall back to the same height that I threw it. In your situation, ground level.)

EDIT:
Or (a fourth option) will the answer depend on the velocity with which I throw it up?

9. Jun 6, 2014

### Nathanael

Let's go back to the original problem.

Pretend that you DO know the time it is in the air. Call it " t "

With that piece of information, do you know how to solve for the displacement?

10. Jun 6, 2014

### Jai

Yes I understand what you mean. Would it depend on the velocity it was thrown at since it will constantly accelerate as it comes back down, meaning that if it has more time to do so it will end up at a greater speed?

11. Jun 6, 2014

### Jai

Yes I could use s=ut+1-2at^2 to calculate that.

12. Jun 6, 2014

### Nathanael

But, it will also be thrown up at a greater speed, too! So does it balance out?

13. Jun 6, 2014

### Jai

Maybe not since it would be decellerating while travelling up and accelerating while coming down?

14. Jun 6, 2014

### Nathanael

So what would "u" be in this situation? What would "a" be?

15. Jun 6, 2014

### Jai

"a" would be 9.8m/s^2 and "u" is 15.6 m/s

16. Jun 6, 2014

### Nathanael

If you're unsure, try using an equation:

If I throw a ball straight up with a known speed, " v " what will be it's maximum height, " h "?

17. Jun 6, 2014

### Nathanael

Is that right?

When you are finding the displacement, you're only finding the horizontal displacement, right? Because the vertical displacement will be zero (it starts at the ground and falls back to the ground)

So when you use this equation, everything will be HORIZONTAL

The equation will be (in words:)

Horizontal displacement = initial horizontal velocity * time + horizontal acceleration * time squared / 2

18. Jun 6, 2014

### Jai

So the time would be equal?

19. Jun 6, 2014

### Jai

I didn't realise that thanks.

20. Jun 6, 2014

### Nathanael

I don't know if you worked it out or if you're just guessing, but yes, the time would be equal (you can find this out mathematically)

So what would "u" and "a" be?

21. Jun 6, 2014

### Nathanael

So if the time it takes to rise and fall are the same, can you think of any way of finding the total time that it's in the air?

22. Jun 6, 2014

### Jai

Well I guess since we are neglecting any vertical forces a is going to equal 0 since there is no air resistance and u will equal the horizontal component which is 15.068.

Last edited: Jun 6, 2014
23. Jun 6, 2014

### Nathanael

Exactly.

Correct

I have no idea if that's true, since you never said what the initial velocity was and you never said what the launch angle was.

Generally though, the horizontal component will equal v*cos(θ) where θ is the launch angle (and v is the initial velocity)

24. Jun 6, 2014

### Nathanael

So we've been beating around the bush a lot (my fault haha) but essentially this is was the problem comes down to:

There are 2 factors involved, the time in the air, and the horizontal speed. If either of these are zero, then the displacement will be zero. (Common sense, right?)
As a matter of fact, the displacement will be equal to the time in the air multiplied by the horizontal speed.
(assuming no air resistance or other horizontal forces)

The time in the air is determined by 1 thing: the vertical speed. (and the gravity, but on Earth that's bascically constant)

The horizontal speed is determined by.... the horizontal speed (obviously haha)

So if you can figure out the time in the air, you can solve the problem (in fact you've already said how to solve it if the time is known)

This is why I've been talking about the time in the air since my first post.

SO ... any ideas on how to solve for the time in the air?

25. Jun 6, 2014

### Jai

None at all unfortunately.