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Projectile Motion of a ball shot into the air

  1. Oct 25, 2006 #1
    A ball is shot from the ground into the air. At a height of 9.1 m, the velocity is observed to be v = (7.9 i + 6.9 j) m/s, with i horizontal and j upward.

    (a) To what maximum height does the ball rise?
    11.5136 m
    (b) What total horizontal distance does the ball travel?
    m
    (c) What is the magnitude of the ball's velocity just before it hits the gound?
    m/s
    (d) What is the direction of the ball's velocity just before it hits the ground?
    ° relative to the horizontal

    I got A

    I don't really know how to start B.. could someone give me a hint?
     
  2. jcsd
  3. Oct 25, 2006 #2

    Doc Al

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    How much time does it spend in the air?
     
  4. Oct 25, 2006 #3
    3.06s?

    Using h = 1/2gt^2 and 11.5136 as the H
     
  5. Oct 25, 2006 #4

    Doc Al

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    OK. Now make use of that to find the horizontal distance traveled.
     
  6. Oct 25, 2006 #5
    \Delta X = Vox*t

    Is that the formula I should use?
     
  7. Oct 25, 2006 #6

    Doc Al

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    That will work.
     
  8. Oct 25, 2006 #7
    What's Vox? If it's 0 it won't work.
     
  9. Oct 25, 2006 #8

    radou

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    The velocity in the x-direction remains constant. You only have to read it off from your velocity vector.
     
  10. Oct 25, 2006 #9
    Alright I did \Delta X = 7.9(3.06) and I got 24.174 which is correct.

    It's amazing how you guys know this stuff, just like right off the top of your head.
     
  11. Oct 25, 2006 #10
    So for C how would i start doc?
     
  12. Oct 25, 2006 #11

    radou

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    Write down the equation for the y-component of velocity, that's a good start.
     
  13. Oct 25, 2006 #12
    V(y) = Vosin(theta)-gt

    Correcto?
     
  14. Oct 25, 2006 #13

    radou

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    Correcto. Now, you can find Vo*sin(theta) = Voy directly from y(t) = Voy*t - 1/2*g*t^2, since, at the total time of the flight you calculated, y has to equal zero. When you do that, use the equation for V(y) you wrote down to claculate the y component of the velocity just before the object hits the ground.
     
  15. Oct 25, 2006 #14
    Well I hope I didn't do it wrong, I got Voy = 14.994 .. but I don't know theta?
     
  16. Oct 25, 2006 #15

    radou

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    Read more carefully. We already said that Voy = Vo*sin(theta), so you don't have to know the angle. :smile:
     
  17. Oct 25, 2006 #16
    V(y) = 14.994 - 9.8(3.06)

    Tried it, it doesnt work :(.
     
  18. Oct 25, 2006 #17

    radou

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    Of course it doesn't work, since you're asked for the magnitude of the velocity, which means the magnitude of the velocity vector. So, you know Vx already. Now, just plug the result you got on V(y) into the velocity vector and find it's magnitude.
     
  19. Oct 25, 2006 #18
    Could you try saying that in a different way? I don't understand what that means..

    Also check your PMs.
     
  20. Oct 25, 2006 #19

    radou

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    The velocity was given with v = (7.9 i + 6.9 j) = (Vx i + Vy j)m/s, as you wrote in the beginning. Now, as said, since Vx is constant, it equals 7.9 at any moment. Plug in the Vy you calculated (the one you thought was wrong) and you'll get v = (7.9 i + Vy j) m/s. The magnitude of a vector is given with the square root of (7.9^2 + Vy^2).
     
  21. Oct 25, 2006 #20
    Sweet it worked! Thanks!
     
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