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Projectile Motion- How to Calculate Displacement

  1. Jun 6, 2014 #1

    Jai

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    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. jcsd
  3. Jun 6, 2014 #2

    Nathanael

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    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?
     
  4. Jun 6, 2014 #3

    Jai

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    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.
     
  5. Jun 6, 2014 #4

    Nathanael

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    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)
     
  6. Jun 6, 2014 #5

    Jai

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    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...
     
  7. Jun 6, 2014 #6

    Nathanael

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    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?
     
  8. Jun 6, 2014 #7

    Jai

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    Yes, it is launched from ground level. Won't the differences in time depend on the angle?
     
  9. Jun 6, 2014 #8

    Nathanael

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    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?
     
  10. Jun 6, 2014 #9

    Nathanael

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    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?
     
  11. Jun 6, 2014 #10

    Jai

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    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?
     
  12. Jun 6, 2014 #11

    Jai

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    Yes I could use s=ut+1-2at^2 to calculate that.
     
  13. Jun 6, 2014 #12

    Nathanael

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    But, it will also be thrown up at a greater speed, too! So does it balance out?
     
  14. Jun 6, 2014 #13

    Jai

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    Maybe not since it would be decellerating while travelling up and accelerating while coming down?
     
  15. Jun 6, 2014 #14

    Nathanael

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    So what would "u" be in this situation? What would "a" be?
     
  16. Jun 6, 2014 #15

    Jai

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    "a" would be 9.8m/s^2 and "u" is 15.6 m/s
     
  17. Jun 6, 2014 #16

    Nathanael

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    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 "?
     
  18. Jun 6, 2014 #17

    Nathanael

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    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
     
  19. Jun 6, 2014 #18

    Jai

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    So the time would be equal?
     
  20. Jun 6, 2014 #19

    Jai

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    I didn't realise that thanks.
     
  21. Jun 6, 2014 #20

    Nathanael

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    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?
     
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