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Projectile motion

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Tom throws a ball of a cliff 45m nigh, with a velocity of 30 m/s. Take accleration due to gravity as 10m/s. (It took 4.5s to reach the bottom)

    What was its horizontal and verticle velocitys when it hit the ground?

    i used the first equation (v=u+at) and with substituion i got -42m/s for the verticle velocity. Is this correct, and how would i find the horizontal? thanks for the help
     
  2. jcsd
  3. Feb 24, 2009 #2

    LowlyPion

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    Welcome to PF.

    If your initial velocity was horizontal, then the time to fall is not given by the velocity relationship. That equation would be useful if you wanted to know how long it took to reach its height i.e when velocity goes to 0.

    To find the time you need to use the relationship that relates distance and acceleration and time.

    x = 1/2*g*t²
     
  4. Feb 24, 2009 #3
    the g stands for gravity right? if so would the answer for the verticle be 101.25?
     
  5. Feb 24, 2009 #4
    and the horizontal 30m/s as its a constant?
     
  6. Feb 24, 2009 #5

    LowlyPion

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    Based on what?

    You'll have to show where that comes from.
     
  7. Feb 24, 2009 #6
    in your equation x = 1/2*g*t²

    So using that i did (assuming the G did stand for gravity). 1/2 * 10m/s*4.5s^2 and i got an answer of 101.25
     
  8. Feb 24, 2009 #7

    LowlyPion

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    You've run out the corral without the saddle there.

    The height is given as 45 m. Your time calculation as I said already is not based on the first equation you applied. 4.5 sec is just plain wrong.

    So start again with the right equation and find the correct time and then you can figure it out.
     
  9. Feb 24, 2009 #8
    Ok using the s=1/2*(u+v)*t i did 45=15t, t=45/15 and so i got 3seconds. Is that correct?
     
  10. Feb 24, 2009 #9

    LowlyPion

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    3 seconds is a much better value to use.
     
  11. Feb 24, 2009 #10
    so what eqution do i use now? :S
     
    Last edited: Feb 24, 2009
  12. Feb 24, 2009 #11
    ok so if the verticle velocity increases by 10 every second, would the answer then be 30m/s?
     
  13. Feb 24, 2009 #12

    LowlyPion

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    Yes the vertical is 30 m/s at impact. As is the horizontal incidentally.
     
  14. Feb 24, 2009 #13
    ah thank you very much. So would that mean the resultant of these would be 54.08? I got that using phythag sqrt 45^2+30^2 = 54.08 with an angle of 34
     
    Last edited: Feb 24, 2009
  15. Feb 24, 2009 #14

    LowlyPion

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    Why do you think it's 452 again?
     
  16. Feb 24, 2009 #15
    http://img21.imageshack.us/img21/7229/83881376.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  17. Feb 24, 2009 #16
    ohhh no it should be 45 it should also be 30! So with that now would the resultant be 42.2, with an angle of 45?
     
  18. Feb 24, 2009 #17
    http://img24.imageshack.us/img24/9337/71116538.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  19. Feb 24, 2009 #18

    LowlyPion

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    Yes. That would be better now.
     
  20. Feb 24, 2009 #19
    Ah yay, thanks for all your help
     
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