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Initial Speed Question

  1. Oct 11, 2006 #1
    At the 18th green of the U. S. Open you need to make a 22.0 ft putt to win the tournament. When you hit the ball, giving it an initial speed of 1.27 m/s, it stops 6.00 ft short of the hole.


    (a) Assuming the deceleration caused by the grass is constant, what should the initial speed have been to just make the putt?
    m/s


    (b) What initial speed do you need to just make the remaining 6.00 ft putt?
    m/s

    So far, to try and find the inital speed needed to make the putt, I have done the following.

    1. 22 feet away from the hole, the putt is 6 feet short, so I subract 6 from 22, and I get 16 feet.

    2. Now that I have 16 feet, I take the initial speed of 1.27 m/s and convert it to feet per second. I get 4.166666673 feet per second.

    3. Now that both the feet away from the hole (16 feet) and the initial speed (4.166666673 ft/s) are in the same units, I divide them, to get the time it took for the golf ball to travel that distance. I get 3.839999994 or 3.84.

    4. Now I am stumped, with these things I have figured out about the problem, how can I use this to find the inital speed I would need to just make the putt?

    Time it takes golf ball to travel 16 feet away from the hole (3.84 seconds)
    Total Distance = 22 Feet
    Golf Ball = 6 feet short from hole

    as for question (b) I am completly stumped. Any help would be greatly appreciated! Thank you!
     
  2. jcsd
  3. Oct 11, 2006 #2

    berkeman

    User Avatar

    Staff: Mentor

    I don't think your step -3- is correct. The velocity is not constant, the deceleration is. Use the kinematic acceleration equation to tell you what the deceleration is in m/s^2.
     
  4. Oct 11, 2006 #3
    the correct formula is Change in Velocity / Time correct? If So I have taken the change in velocity, I have taken 1.27 m/s and divided it by 3.84(the time taken for the ball to reach the hole)

    This gives me the number .330729167 or, .3301

    When I plug this into a position versus time graph, I get

    xf=xi+viT+.5at^2

    0=1.27+0+.5(.3301)t^2


    0=1.27+.1653645833 or

    0=1.27+.17t^2

    now I subtract

    0=1.27+.17t^2
    -1.27

    -1.27=.17t^2

    t^2=-.1338582677 (or -.134)

    but I cannot get the square root of -.134 on my calculator because it says ERROR: NON REAL ANSWER, so I am not sure where I messed up, but any help is still appreciated!
     
  5. Oct 11, 2006 #4
    ****Bump****
     
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