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Cannon Problem (Projectile motion)

  1. Dec 26, 2008 #1
    1. The problem statement, all variables and given/known data
    An Olympian in the long-jump goes into the jump with a speed of 11 m/s at an angle of 20 (degrees) above the horizontal. How far does the Olympian jump?


    2. Relevant equations
    Constant Acceleration Equations
    V=Vo + at


    3. The attempt at a solution
    I tried to set up a X/Y Chart to solve for V, Vo, Displacement in X and Y, Acceleration, and Time. However. I'm confused on how to solve for V in the Y direction. I know that Vo and V in the X direction are equal, but how about in the Y direction? How do you solve for that? And how do you solve acceleration in the Y direction? Thanks.
     
  2. jcsd
  3. Dec 26, 2008 #2

    mgb_phys

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    In the y (vertical direction) there is an accelration downwards of g = 9.81m/s^2
    You can also use the fact that the vertical velocity is zero at the top of the jumpand this is halfway through the total time
     
  4. Dec 26, 2008 #3
    But isn't F=mg? How can you assume that the mass is 1? I'm studying for my final exam so I kind of forgot all of this. : (
     
  5. Dec 26, 2008 #4

    jgens

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    Alright, your first step in this problem should be to resolve the velocity into its respective x and y components. Do that first and go from there.

    Edit: mgb_phys never assumed the mass of jumper was 1 kg. Assuming the jumper is on earth, the acceleration in the y component is 9.81 m/s^2. Use the fact that the velocity in the y component is zero halfway through the jump to calculate the time in the air.
     
  6. Dec 26, 2008 #5
    How do you solve for V in the y direction though? I know Vo and V in X are the same...I found out Vo in Y..I can't figure out V in Y though.
     
  7. Dec 26, 2008 #6

    jgens

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    Show some of your work so we can see what you're doing.

    Assuming you correctly calculated the initial vertical velocity of the jumper, use the formula V_f = V_i + at.
     
  8. Dec 26, 2008 #7
    But I can't solve for T yet. How should I solve for T? This is where I get stuck.
     
  9. Dec 26, 2008 #8

    jgens

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    You've already been told twice. Use the fact that the velocity in the y component is zero half way through the jump to calculate the time.
     
  10. Dec 26, 2008 #9
    So I use T/2 and V=0 and use a constant acceleration equation? Thanks for helping me btw
     
  11. Dec 26, 2008 #10

    mgb_phys

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    Work out the vertical component of the velocity
    you have v = vo + at
    Consider just the upward half of the curve, you know the initial and final velocity and the 'a' so t should be easy - remember the time to the top is half the total time.
     
  12. Dec 26, 2008 #11
    Thanks for helping me. That last part did the trick : )
     
  13. Dec 26, 2008 #12
    You have the launch velocity (Vlaunch = 11 m/s) and the launch trajectory (20 degrees above horizontal). Now you need to derive the following:

    Vertical velocity component (Vvert, in m/s):

    Vvert = Vlaunch x SIN(20 degrees)

    --------------------------------------------------------

    Vertical rise time (Vrt, in seconds):

    Vrt = Vvert / (9.8 m/s^2)

    ---------------------------------------------------------

    Vertical height achieved (Vh, in meters):

    Vh = ½ (9.8 m/s^2) time^2

    ---------------------------------------------------------

    Horizontal velocity component (Vhor, in m/s):

    Vhor = Vlaunch x COS(20 degrees)

    ---------------------------------------------------------

    Horizontal distance = Vhor x 2Vrt

    Technically speaking, you’re scenario is missing one initial factor; that being the height in which the javelin is actually being released from, as that adds to the rise and fall time thereby yielding just slightly greater distance in both the horizontal and vertical. However, this aspect is simply overlooked in many cases for such low initial launch heights (the javelin potentially being released from a height of only 4.5 to 6 feet). The fact that you weren’t given the javelin’s release height means it’s likely not an issue, but it’s good to realize it actually is a consideration in yielding the greatest calculated accuracy.
     
    Last edited: Dec 26, 2008
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