Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?

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

    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


    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



    0=1.27+.1653645833 or


    now I subtract



    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook