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Projectile; Golf example, Please help!

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data

    I know this is a basic question but this is my first physics course ever and I am having big time trouble, any help is appreciated!

    Chipping from the rough, a golfer sends the ball over a 3.00 m high tree that is 14.0 m away. The ball lands at the same level from which it was struck after traveling a horizontal distance of 17.8 m - on the green of course! (A) If the ball left the club 54.0 degree angle above the horizontal and landed on the green 2.24 s later, what was its initial speed? (B) How high was the ball when it passed over the tree?


    I have found tons of examples but they seem to ALL have the initial velocity as a given, and I think I have thoroughly confused myself with so many equations today.
     
  2. jcsd
  3. Jul 6, 2008 #2
    what have you tried so far? what are the equations you are using?
     
  4. Jul 6, 2008 #3
    I have been trying to split it up into its horizontal and vertical motions, using some equations like Vf = (Vi)2a(delta)d and d = (Vi)t + 1/2at2 but to be honest I'm not too sure what I'm doing with them... just seems to be too many unknowns, and I'm not particularly strong in trig.
     
  5. Jul 6, 2008 #4
    I have:

    2.24 s = (delta)V / a

    2.24 s = (delta)V / -9.80 m/s2

    (2.24 s)(-9.80 m/s2) = (delta)V

    (delta)V = -22 m/s

    Vi = 11 m/s ; but I am not sure of anyyy of this :(
     
  6. Jul 6, 2008 #5
    The horizontal aspect I think is fairly straightforward, using VH = d / t;

    VH = 17.8 m / 2.24 s for 7.95 m/s ...

    I am still lost on finding the initial vertical(y) velocity however..
     
  7. Jul 6, 2008 #6

    If you know that the Vx= 7.95 m/s:

    Vx=Vo*cos(angle), you know the angle, so solve for Vo.

    Then you can use Vy(horizontal)= Vo*sin(angle)-(9.8m/s^2)(time) to get the vertical velocity.
     
  8. Jul 7, 2008 #7
    There is probably several ways to solve part "B". One way is the remember that projectiles are inverted parabolas (axis of symmetry parallel to the y-axis, opens down) of the form:

    (x-h)^2= -4a(y-k) where (h,k) is the vertex of the parabola. Use equations of projectile motion to determine the location maximum height. Pts (0,0) and (17.8,0) lie on the parabola so you can determine "a". Since you know the tree lies at x= 14 you can find the height of the parabola at that point and compare it to the tree height.
     
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