Projectile Motion- How to Calculate Displacement

[Jai]

Homework Statement

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.

The Attempt at a Solution

first attempted to approach this by calculating the horizontal and Vertical initial velocity. I then realized 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

 

[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?

 

[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.

 

[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)

 

[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...

 

[Nathanael]

the initial vertical velocity is 4.038 m/s^2

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?

 

[Jai]

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

 

[Nathanael]

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

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?

 

[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?

 

[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?

 

[Jai]

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

 

[Nathanael]

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?

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

 

[Jai]

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

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

 

[Nathanael]

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

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

 

[Jai]

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

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

 

[Nathanael]

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

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 "?

 

[Nathanael]

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

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

 

[Jai]

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 "?

So the time would be equal?

 

[Jai]

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

I didn't realize that thanks.

 

[Nathanael]

So the time would be equal?

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)

I didn't realize that thanks.

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

 

[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?

 

[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.

 

[Nathanael]

Well I guess since we a re neglecting any vertical forces a is going to equal 0 since there is no air resistance.

Exactly.

u will equal the horizontal component

Correct

which is 15.068.

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)

 

[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?

 

[Jai]

None at all unfortunately.

 

[Nathanael]

None at all unfortunately.

Do you know how to solve for the time it takes to reach it's maximum height?

 

[Jai]

Do you know how to solve for the time it takes to reach it's maximum height?

No I don't.

 

[Nathanael]

No I don't.

acceleration * time = change in velocity

therefore:

time = change in velocity / acceleration


Also:

Change in velocity = Final velocity - initial velocity



So do you know the initial velocity, final velocity, and acceleration?

 

[Jai]

Oh wow. Just realized something... 15.6 m/s is the velocity and 0 is the initial velocity... Sorry about that...

 

[Nathanael]

15.6 m/s is the velocity and 0 is the initial velocity

0 is the initial velocity?

 

[Jai]

Yes, since a bullet is being shot out of a gun.

 

[Nathanael]

Yes, since a bullet is being shot out of a gun.

If the velocity is zero, then it's not being shot out of a gun, it's just sitting in a gun.


When I said "initial velocity" in my other post I meant "initial velocity" after it's been shot

 

[Jai]

If the velocity is zero, then it's not being shot out of a gun, it's just sitting in a gun.

Yes, however I mean that for my original question, so now the approach may be different?

 

[NascentOxygen]

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.

Those equations are good. They are what you use. You just need to apply them separately to the horizontal component of initial velocity and to the vertical component of initial velocity. Vertical motion is influenced by gravity, horizontal motion is influenced by only air resistance (if present).

If a projectile has no initial horizontal velocity component then its motion will just be a vertical rise, then fall back to where it started.

 

[Jai]

Those equations are good. They are what you use. You just need to apply them separately to the horizontal component of initial velocity and to the vertical component of initial velocity. Vertical motion is influenced by gravity, horizontal motion is influenced by only air resistance (if present).

If a projectile has no initial horizontal velocity component then its motion will just be a vertical rise, then fall back to where it started.

Thanks, I applied them separately and managed to get what I need. While you are here do you know some sort of formula that could be used to take into account air resistance? My teacher said that we need to mention the formula but not calculate it because it is too complicated to get the data for.