Projectile Motion- How to Calculate Displacement

Click For Summary

Homework Help Overview

The discussion revolves around calculating the displacement of a projectile launched at various angles, given the initial velocity and acceleration due to gravity. Participants explore the relationships between vertical and horizontal components of motion, as well as the effects of launch angle on time of flight.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss calculating initial horizontal and vertical velocities and question the impact of air resistance. There are attempts to apply kinematic equations to determine time in the air and displacement, with some uncertainty about the correct application of these equations.

Discussion Status

The conversation is ongoing, with participants offering guidance on how to approach the problem. There is a focus on understanding the relationship between time of flight and horizontal displacement, but no consensus has been reached on specific calculations or methods.

Contextual Notes

Participants are working under the assumption that air resistance is negligible and that the projectile is launched from ground level. There is some confusion regarding the definitions of initial velocity and acceleration, as well as the correct application of kinematic equations.

Jai
Messages
17
Reaction score
0

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
 
Last edited:
Physics news on Phys.org
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?
 
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.
 
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)
 
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...
 
Jai said:
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?
 
Yes, it is launched from ground level. Won't the differences in time depend on the angle?
 
Jai said:
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?
 
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
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
Yes I could use s=ut+1-2at^2 to calculate that.
 
  • #12
Jai said:
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?
 
  • #13
Nathanael said:
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?
 
  • #14
Jai said:
I could use s=ut+1-2at^2 to calculate that.

So what would "u" be in this situation? What would "a" be?
 
  • #15
Nathanael said:
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
 
  • #16
Jai said:
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 "?
 
  • #17
Jai said:
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
 
  • #18
Nathanael said:
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?
 
  • #19
Nathanael said:
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.
 
  • #20
Jai said:
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)

Jai said:
I didn't realize that thanks.
So what would "u" and "a" be?
 
  • #21
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
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:
  • #23
Jai said:
Well I guess since we a re neglecting any vertical forces a is going to equal 0 since there is no air resistance.
Exactly.

Jai said:
u will equal the horizontal component
Correct

Jai said:
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)
 
  • #24
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
None at all unfortunately.
 
  • #26
Jai said:
None at all unfortunately.

Do you know how to solve for the time it takes to reach it's maximum height?
 
  • #27
Nathanael said:
Do you know how to solve for the time it takes to reach it's maximum height?

No I don't.
 
  • #28
Jai said:
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?
 
  • #29
Oh wow. Just realized something... 15.6 m/s is the velocity and 0 is the initial velocity... Sorry about that...
 
  • #30
Jai said:
15.6 m/s is the velocity and 0 is the initial velocity

0 is the initial velocity?
 

Similar threads

Find the height of a lamp based on the velocities of projectiles
  • · Replies 40 ·
2
Replies
40
Views
3K
Calculate the displacement? (Projectile motion problem)
  • · Replies 3 ·
Replies
3
Views
2K
Find Projectile Flight Time Given Only Maximum Height
  • · Replies 4 ·
Replies
4
Views
3K
I need opinions on a Projectile Motion problem that I made up
  • · Replies 11 ·
Replies
11
Views
2K
Launching a Projectile with Air Resistance
  • · Replies 13 ·
Replies
13
Views
3K
Relationship between horizontal range and angle of launch
  • · Replies 7 ·
Replies
7
Views
4K
Projectile Motion Experimental Error
  • · Replies 1 ·
Replies
1
Views
2K
Projectile Motion : Work and Energy
  • · Replies 4 ·
Replies
4
Views
2K
Projectile Motion on the Moon: Finding the Optimal Launch Angle Using Equations
  • · Replies 2 ·
Replies
2
Views
3K
Relationship between Launch height and range of a projectile
  • · Replies 15 ·
Replies
15
Views
27K