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Need help on projectile motion equation

  1. Feb 12, 2008 #1
    Need help on projectile motion equation!!!

    okay today i dazed off in class and didn't really listen to how my teach told me how to do projectile motion equations. here's the one i'm working on now.

    A quaterback tries to throw a football to his receiver at an angle of 30 degrees with respect to the ground. The receiver is at a distance 30 meters from the quarterback.

    (a) The receiver stands still. At what speed must the quarterback throw the ball so the receiver can catch it? (Assume that the paint of release and the point of reception are at the same height)

    i've already split the x and y components apart and wrote down what i know. other than that im seriously stuck.

    Would draw a picture for you but cant really :) plz somebody get back to me
  2. jcsd
  3. Feb 12, 2008 #2
    What do you know about the distance (horizontally) that the ball has travelled when it reaches its maximum height?
  4. Feb 12, 2008 #3
    idk if i were to guess it'd be half the x component wouldnt it
  5. Feb 12, 2008 #4


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    Vertically, the speed would be Vsin30, right?
    since acceleration=g which is constant, the kinematics equations apply

    so if [itex]v=u+2at[/itex] where v=final vertical speed,u=initial vertical speed

    When the ball reaches the maximum height, what would be its final velocity?(Hint:As the ball goes up, the final velocity decreases)
  6. Feb 12, 2008 #5
    why would the vertical be vsin30?
  7. Feb 12, 2008 #6


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    The speed at 30 degrees is V. When you spit that into the x and y components, the initial vertical velocity is Vsin30
  8. Feb 12, 2008 #7
    oh okay that makes sense now.
    working it out now
  9. Feb 12, 2008 #8
    how does this help me find the initial velocity of the ball
  10. Feb 12, 2008 #9


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    The Vsin30 is the initial velocity.At the maximum height what do you think the final velocity is?
    Last edited: Feb 12, 2008
  11. Feb 12, 2008 #10
    idk....you gotta kinda spoon feed me here. lol
  12. Feb 12, 2008 #11


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    Well as the ball rises..the vertical velocity decreases...until it reaches the max height. After reaching the maximum height, the vertical velocity begins to increase...

    the only for that to happen is if the velocity at the max height is Zero.

    and the time taken to reach the maximum height is the same as the time taken to go from the max height to the ground.

    (Note: this is just one method, another kinematic equation can be used to find the time taken to reach the receiver as well)
  13. Feb 12, 2008 #12
    have u got the answer cuz I've gotten 26.1 m/s
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