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Projectile prblem

  1. Sep 17, 2008 #1
    I have a question about a projectile motion problem.

    So the background info of this problems is that a quarterback is throwing a football to a receiver who's running with constant v_r and he's now D distance away from the quarterback. The quarterback thinks that it should be thrown at angle theta and the receiver must catch the ball at t_c. The ball is thrown and caught at the same height.

    Now, I have to find v_0y in terms of t_c and g
    I just thought it was vsintheta.... how do you find it that way? Is it v = 1/2gt^2??
    and the v_0x, I have to find in terms of D, t_c, and v_r.....
    I am totally lost here... please help!
     
  2. jcsd
  3. Sep 18, 2008 #2

    rl.bhat

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    Whether the receiver is approaching to or receding from the quarterback?
     
  4. Sep 18, 2008 #3
    Lets break the problem up a bit.

    First work out:

    {1} How far away the receiver will be when the ball arrives at his location at time tc

    {2} If the quarterback throws the ball at t=0 and its caught at tc what horizontal speed must it have to travel the distance worked out in {1} to get to the receiver at time tc.
     
  5. Sep 18, 2008 #4
     
  6. Sep 18, 2008 #5
    1) the receiver's distance would be d = v_rt_c??
    2) v_0 = d / t_c????

    i don't get how these help us find the component vectors of v_0...
     
  7. Sep 18, 2008 #6
    The start point of the receiver is not entirely clear. But I understand that he does not start right next to the quarterback, that is, there is a distance D between them.

    Therefore the ball must travel a horizontal distance x = D + V_r*tc

    The horizontal velocity of the ball then must be x/tc.

    Hint... if we assume that the two guys playing football are the same height then you should take the total time the ball is in the air (which is tc) and calculate how much vertical velocity you would have to throw a ball straight up with to have stay in the air (above the height of the footballers hands) for time tc.

    Once you have this you can work out the angle as you will have Vy and Vx... (use tan(theta))
     
  8. Sep 18, 2008 #7
    so for v_x, in terms of D, t_c, and v_r, it would be
    D + v_rt_c / t_c

    for v_y....
    do I use the equation, yt_c = v_1t_c +1/2at_c^2??
    and then
    v_y = yt_c - 1/2at_c^2?
    but it says to express in terms of t_c and g. . . . . . . but I have y in the equation, referring to the height..
     
  9. Sep 18, 2008 #8
    i'm so confused..... T.T
     
  10. Sep 19, 2008 #9
    Okay... so you need the answer expressed in terms of tc and g.

    The ball is thrown up in the air and takes a time tc to return to the original height. At the heighest point of the balls trajectory the the vertical velocity will be zero and the time will be tc/2.

    Therefore.... using: Vv=Uv+at

    0= Uv - 9.8*tc/2

    Uv = 9.8*tc/2 (Uv being the initial vertical velocity)

    The Ux (initial horizontal velocity) will be found by simply using.. Speed = distance/time

    Ux = (D+Vr*tc)/tc (Vr being the velocity of the guy running for the ball).

    I hope this helps, if not.. let me know.
     
  11. Sep 19, 2008 #10
    ok now i understand how to get the horizontal velocity
    but how would you do v_y??
    you said to consider the velocity needed for the ball to stay up...
    you don't know how far the ball goes up vertically... so how do you figure it out?
     
  12. Sep 21, 2008 #11
    What I called Uy, you have called V_y.

    When you throw an object up in the air, the more initial speed you give it the more time it will spend up in the air before it reaches its highest point.

    We know in this case what the total time for the ball to be in the air is, that is the amount of time it takes to travel between the guy throwing and the guy catching. Half this time is the amount of time for the ball to reach maximum height. Think of the trajectory of an object being thrown.

    I assume you are familiar with the expression:

    V = U + at

    where V is the final velocity, U is the initial velocity, a is the acceleration, and t is the time.

    Well, midway in the trajectory of the ball we V=0 as it has reached its maximum height. We know what the acceleration is (g=-9.81) and we know what the time at this point is (half the total time).

    To calculate the balls initial velocity we do not need to know how far it goes up, but we have enough information here to work that out though.
     
  13. Sep 21, 2008 #12
    ok... so i found v_y to be 1/2gt and v_x to be D + v_rt/t
    now... it says to find the speed v_0 with which the quarterback must throw the ball in terms of D, t_c, v_r, and g...
    so i did sqaure root of ((1/2gt_c)^2 + (D+v_rt_c/t_c)^2)
    and it says it's wrong....why???????
     
  14. Sep 21, 2008 #13
    to find v_0, weren't u just supposed to add the vectors together??? either i'm totally misunderstanding the concept.. or this question's weird lol
     
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