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Projectile Motion golf ball

  1. Sep 23, 2009 #1
    Hello, im having trouble with the question below..

    1. A golfer imparts a speed of 30.3 m/s to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. A) How much time does the ball spend in the air? b) what is the longest "hole in one" the golfer can make if the ball doesn't roll?



    2. right triangle, T=(Vo-V)/g, for b i know the eqn is x= Vx*t


    3. i made a right triangle with the right angle on the bottom right. i made the hypotenuse 30.3 m/s. I believe the equation to find time is T=(Vo-V)/g. How would i find the time?

    thank you
     
  2. jcsd
  3. Sep 23, 2009 #2

    rock.freak667

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    use

    [tex]y=y_0 +v_y t -\frac{1}{2}gt^2[/tex]


    at what angle must the golfer hit the ball to get the maximum range?
     
  4. Sep 23, 2009 #3
    it doesn't give an angle, would y(the height) be 0 since it is not changing elevation?
     
  5. Sep 23, 2009 #4

    rock.freak667

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    yes it would so your equation is y=vyt-(1/2)gt2 and you want to find when y=0

    also it said it hit it to the maximum range, so there is only one angle the max range can be achieved, do you know what this angle is?
     
  6. Sep 23, 2009 #5
    the only angle it gives is 90°, so 90?, i got 90 from the triangle that i drew..
    the problem that i keep running into is that i get 0=30.3 m/s(t) - (1/2)(-10 m/s^2)(t^2)..... and that gives me t=0
     
    Last edited: Sep 23, 2009
  7. Sep 23, 2009 #6

    rock.freak667

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    if the range is given by

    [tex]R=\frac{v^2 sin 2\theta}{g}[/tex]

    for what angle is R maximum for?
     
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