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Help with a problem

  1. Oct 3, 2003 #1
    Help with a problem!!

    This problem seems pretty easy but for the life of me I cannot get it, I think it's confusing me because I don't have a intial velocity or a time...help!

    A golfer hurls his golf club across the green. Launched at a angle 30 degrees above the horizontal, the club lands 50 m away, what was the launch speed of the 4 kg club?

    I've tried this problem over and over and I can't get it, as far as I can tell the mass is useless right?

    The way I tried it before was by using one of the kinematic equations, I now know that this problem has to be split into it's x and y components, from there...I need help.
  2. jcsd
  3. Oct 3, 2003 #2


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    Yes, mass is unimportant.

    Make an expression for x velocity relating time spent in flight and how far the club flew.

    Make an expression for initial y velocity using time of flight and gravitaional acceleration.

    Relate the expressions for x velocity and y velocity, eliminating time, using the launch angle.

  4. Oct 3, 2003 #3
    Perhaps this will help, using a range formula.

    We will assume no air resistance and be using g as the accleration due to gravity and sin Θ to represent the heigth of the clubs in the air (2Θ since the clubs must go up and down) we achieve the following variant of a kinematic equation:

    x = Vinitial squared * sin(2Θ)/g. Solving for Vinitial = square root of gx/sin(2Θ) or 31.3 m/s

    The final answer is the actual initial velocity not just the x-component.

    NOTE: The range formula only works if the initial take off point and the landing point are of the same height. As the initial height of the man is unknown, you will have to assume the clubs land at the same heighth as the man (on a hill perhaps) or the man is laying on the ground when they are thrown.
    Last edited: Oct 3, 2003
  5. Oct 3, 2003 #4


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    An object moving with a constant horizontal speed vx has distance function x= x0+ vx t.

    An object moving with initial vertical speed vy and constant vertical acceleration -g has distance function y= y0+ vy t- (1/2)g t2.

    In this problem you can take x0 and y0 to be 0. The club hits the ground when y= 0 (again) so that gives you t. Put that into
    x= x0 + vx t= 50 to find the initial speed.
  6. Oct 4, 2003 #5
    Game Guru , I am correcting a test I have already had so I have it to study, and your answer is what I put on the test and got a 0/15...so it's not that.

    This is what I'm thinking,
    I use the Range formula:
    range=initial velocity squared divided by g, and that multiplied by sin 2 theta,

    or 50=(Vo^2/9.8) *sin 2(30 degrees)which gives me 23.8 m/s...Am I close? Anybody?

    I still can't figure out how to do this when all the equations leave me which both time and initial velocity missing.
  7. Oct 4, 2003 #6
    My apologies ConfusedStudent. Although the range formula is correct, my initial calculation may have been done in error. When double-checking today, I do get 23.8 m/s as V0.

    Here is another way of solving the problem.

    The horizontal displacement of the clubs is given by x=V0*t+at2/2 with a = 0: x=V0 * t = (V0 cosθ)t. The vertical component of the club's velocity at any time t is given by Vy = Vinitialy + ayt. At the instant that the clubs land, vy=-v0y. Therefore, the vertical component of the club's velocity is

    -v0y=v0y + ayt

    If you solve for t (up being positive):

    t = -2voy/ay = -2v0sinθ/ay

    Plug this expression for t into the equation for x gives you

    x=(v0cosθ)t = v0cosθ(-2v0sinθ/ay)

    or in more simpler terms:


    Solving for V0 = Square root of -xay/(2cosθsinθ) = 23.8 m/s.

    Hope that makes it clearer and sorry for my late night miscalculation.
    Last edited: Oct 4, 2003
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