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Homework Help: Projectile Motion (Displacement and Velocity vector)

  1. Sep 20, 2005 #1
    hey guys, i have this quesiton im diong, i got allt he answers but theres one prob

    Question:

    A projectile is fired at 50 m/s and 60 degrees along a level surface.

    a) Find the velocity and displacement VECTORS at t = 1.0 s

    i used v2 = v1 + at
    v2 = 43.3 + (-9.8) (1)
    v2 = 33.5 m/s
    thats the velocity at 1 second

    now displacement i used
    d = v1t + 1/2 at squared
    i got y = 43.3 (1) + 1/2 -10 (1) squared
    to get
    displacement = 38.4 meters

    ofcourse the y velocity i got it from : 50 sin 60 = 43.3 m/s and x velocity 50cos60 = 25 m/s as you know

    but teh teacer said those aer right numbers but he ased for VECTORS so he wants 33.5 and an angle beside it i assume, i am lost now, i need help finding the displacement and velocity VECTORS :D
    thanks guys!

    ---
    i posted this in the other forum too , didnt know which ones more right lol thanks, but need it soon
     
  2. jcsd
  3. Sep 20, 2005 #2

    Doc Al

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    Staff: Mentor

    That's just the y-component of the velocity. You need to combine this with the x-component to get the correct velocity vector.

    Similar comments apply for displacement.
     
  4. Sep 20, 2005 #3
    ok thats what i dont know how to do :( can you give me instructions, thanks brother :)
     
  5. Sep 20, 2005 #4
    actually i think i get it
     
  6. Sep 20, 2005 #5
    got the answers :)

    just checking

    33.5 and 64.6 degrees for velocity

    displacement is 38.4 and 58.4 degree
     
  7. Sep 20, 2005 #6

    Doc Al

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    No. As I said before, the 33.5 is just the y-component of the velocity. What's the x-component? What's the magnitude? What's the angle of the velocity vector?
     
  8. Sep 20, 2005 #7
    i just said 64.6 degrees :D
    so
    33.5 m/s then theres that angle thingy and i put 64.6 degrees in it
    right
     
  9. Sep 20, 2005 #8

    Doc Al

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    Answer the questions in my last post.
     
  10. Sep 20, 2005 #9
    y comp: 33.5
    x comp: 15.2
    magnitude: you mean direction?
    angle: 64.6 degrees

    he said he wants the answer as a vector, can you please make this easier on me and tell me what the forma tof a vector is :D

    os (x,y)?

    or
    33.5 then <64.6 degrees

    or (x,y) then < 64.6 degrees

    got to sleep soon and the asnwer is due 2morrow, thanks!
     
  11. Sep 20, 2005 #10

    Doc Al

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    OK
    No. [itex]v_x = 50 \cos (60)[/itex]
    No, I mean magnitude. If you have the components, use the Pythagorean theorem to find the magnitude.
    This is wrong since your x-comp is wrong.

    One way is to give a magnitude and a direction. (You can also express it as components, but I suggest magnitude and direction.)
     
    Last edited: Sep 21, 2005
  12. Sep 20, 2005 #11
    isn't the X usually cos 60, not sin60?
    also he wants it for the FIRST second, t = 1
    so 50cos60 is initial, thats why i got 15, for t = 1 seconds
     
    Last edited: Sep 20, 2005
  13. Sep 21, 2005 #12

    Doc Al

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    You are right about [itex]v_x = 50 \cos (60)[/itex]--that's what I meant to write. But since [itex]\cos (60) = 0.5[/itex], [itex]v_x = 25[/itex], not 15.
     
  14. Sep 21, 2005 #13
    true
    i got 15 casue i did
    v2 = v1 + at
    so 25 - 10 (1) cause he wants 1 second
    is 15, as the x displacement
    did the same thing with Y and u said its correct
    im confused :(
     
  15. Sep 21, 2005 #14

    Doc Al

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    Only the vertical direction is accelerated (gravity acts down, not sideways). The horizontal speed remains constant.
     
  16. Sep 21, 2005 #15
    goddd no wonder i got 9/10
    'mechanical error'
    lol i wish u told me yesteryda
    thanks bro
    case closed :D
     
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