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Angled Projectile Motion Question

  1. Jul 9, 2008 #1
    1. An object was thrown up at an angle of 22o from the horizontal. If this object travels 38 m horizontally, what was the speed at which the object was thrown?

    angle = 22o
    dhorizontal = 38m
    g = -9.81m/s2


    I have absolutely no clue on how to even go about starting to solve this question...would anyone be able to give me a small hint perhaps on which formula to work with?
     
  2. jcsd
  3. Jul 9, 2008 #2

    rock.freak667

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    Put the intial speed as V inclined at the angle of 22. Split that into vertical and horizontal components. (Note that the horizontal velocity is constant).
    Then use your kinematic equations.
     
  4. Jul 9, 2008 #3
    If I have the angle and split it into the x and y components, wouldn't I still need at least one velocity given to solve for any of the other sides?
     
  5. Jul 9, 2008 #4

    rock.freak667

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    Find the total time for the motion ([itex]s=ut+\frac{1}{2}at^2[/itex]) in terms of V.

    Then the total horizontal distance travelled=Horizontal Speed * Total time for the motion
     
  6. Jul 9, 2008 #5
    Hmm. The formulas I used were just slightly different, I've never seen that one before...what would the variables of "s" and "u" stand for?
     
  7. Jul 9, 2008 #6

    rock.freak667

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    s=displacement
    u=initial velocity
     
  8. Jul 9, 2008 #7
    Okay, thank you.
    And Vi would be...22?
    Sorry for all the questions. I wasn't having any problems in the whole unit up until this last question!
     
  9. Jul 9, 2008 #8

    rock.freak667

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    Vi=initial veritcal velocity=Vsin22.
     
  10. Jul 9, 2008 #9
    Alright...I've gotten it to:

    t = -2(Vsin22o) / -9.81m/s2
     
  11. Jul 9, 2008 #10

    rock.freak667

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    Good.

    Now consider horizontal motion.

    Horizontal Distance=Horizontal speed * Time


    and you have t=(2Vsin22)/9.81 and the horizontal distance is 38
     
  12. Jul 9, 2008 #11
    Okay...So 38 m = V x (2vsin22/-9.81)

    Are you able to combine the two V's, if one is horizontal and one is vertical?

    When I tried that in the next step I have -372.78m=2v2sin 22o...when working through it though, it ends up with the root of a negative number so I'm assuming I made a mistake somwhere. =/
     
  13. Jul 9, 2008 #12

    rock.freak667

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    [tex]\frac{-2(Vsin22)}{ -9.81}[/tex]

    minus signs cancel out.

    and Horizontal velocity=Vcos22.
     
  14. Jul 9, 2008 #13
    Solving for V, I got 23 m/s. According to the book that is the correct answer!

    This will add 3% onto my overall mark :) Thank you, I really owe you one!
     
  15. Jul 10, 2008 #14

    rock.freak667

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    Once you understand how to do the problems is all that's needed.
     
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