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Projectile Motion: golf ball velocity to reach the green

  1. Dec 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A golfer strikes a golf ball at an angle of 17 degrees above the horizontal. With what velocity must the ball be hit to reach the green which is a horizontal distance of 250 m from the golfer? It is at the same height as the tee.

    2. Relevant equations
    trig equations
    v2^2=v1^2+2ad
    d=v1(t)+1/2a(t)2
    d=(v1+v2/2)t
    3. The attempt at a solution
    Well I used trigonometry to find the dy (tan17*250) and I believe that's the maximum height.......meaning v2y at that time is 0
    So I have dy=76.43 m
    a=-9.8m/s^2
    v2y=0
    I used v2^2=v1^2+2ad to find the initial velocity in the y direction and got 38.7
    I plugged that into the equation d=(v1+v2/2)t to find time and got 1.975 seconds
    For the x direction I know dx=250, a=0, and t=1.975 seconds
    So I used d=v1(t)+1/2a(t)2 to find v1 but the a is 0 so the equation just ends up being d=v1(t) and I got the initial velocity in the x direction to be 126.59
    I used pythagorean theorem to find the total velocity to be 132.34 m/s 17 degrees above the horizontal....but the answer is 66m/s 17 degrees above the horizontal.
     
  2. jcsd
  3. Dec 4, 2016 #2

    Dr Transport

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    use the range equation [itex] R = \frac{v_{0} ^{2}sin(2\theta)}{g}[/itex]
     
  4. Dec 4, 2016 #3

    haruspex

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    Wouldn't that be the height by which the ball would fly over the green if there were no gravity?
     
  5. Dec 4, 2016 #4

    haruspex

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    That would certainly be simpler, but I prefer first to help the student fix the errors in the chosen method.
     
  6. Dec 4, 2016 #5
    Sorry I don't understand
     
  7. Dec 4, 2016 #6

    haruspex

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    You have not posted a diagram, but I hope you drew one.
    The launch angle is 17 degrees and the length of the hole 250m. To reach a height of 250 tan 17 it would have to keep going in a straight line (so no gravity) for the entire distance.
     
  8. Dec 4, 2016 #7
    So dy isn't 250 tan 17...it's 0 because it lands on the ground. But then we have nothing to input into the equations?
     
  9. Dec 4, 2016 #8

    haruspex

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    Sure you do. Try it.
     
  10. Dec 4, 2016 #9
    dy:0 dx:250 v2^2=v1^2+2ad ?=?+2(-9.8)(0)
    t:? t:? v2=v1+at ?=?+-9.8?
    a:-9.8 a:0 d=v1(t)+1/2(a)(t)^2 0=?(?)+1/2(-9.8)(?)^2
    vy1:? vx1:? d=v2(t)-1/2(a)(t)^2 0=?(?)-1/2(-9.8)(?)^2
    vy2:? vx2:? d=(v1+v2/2)(t) 0= (?+?/2)(?)
    I don't know what missing information I'm able to get from the two numbers given in the original question? I can't think of anything
     
  11. Dec 4, 2016 #10

    gneill

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    Moderator note: Thread title changed to make it descriptive of the problem. Thread titles should not be overly generic! "Projectile motion" was far too generic.
     
  12. Dec 4, 2016 #11
    whoops sorry
     
  13. Dec 4, 2016 #12

    haruspex

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    Since you have the launch angle, the vertical requirements (d=0) give you a relationship between the initial speed and the flight time.
    The horizontal requirements give you another relationship between them.
    Solve the simultaneous equations.
     
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