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Having problem with this.

  1. Sep 15, 2012 #1

    A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 17.3 m/s at an angle of 50.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

    2. Relevant equations
    v^2 = v0^2 + 2ax

    3. The attempt at a solution
    Basically, what i did was getting the y value from the speed.
    It was like: 17.3sin(50.0) = 13.25m/s. From the given, what
    I could figure out were acceletion (-9.81m/s) and velocity (13.25m/s)
    Since I only knew those two, I thought that I must use v^2 = v0^2 + 2ax.
    For the maximum height, I did 0 = (13.25)^2 + 2(-9.81)(x) and x was 8.948m.
    However, there was 3.00 elevation, so I subtracted 3.00 from 8.948,
    which was 5.948m. From this, I tried to do the same equation and I
    got the 10.80m/s for the answer, but the internet site says that it is wrong.

    2. Relevant equations

    3. The attempt at a solution

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 15, 2012 #2
    Maybe you can do it with conservation of energy method(if you have learned them) .
    Take the green level as zero level of PE.

    The question is about the speed.
    Using kinematic, you have to find both vertical and horizontal component.
    Surely the horizontal component remains constant.
    So it is the resultant of horizontal and vertical component.
  4. Sep 15, 2012 #3
    I got the same answer as you. Intuitively, our answer should be right, because if you calculate for the final velocity of the golf ball when it lands at the same height from which it was hit (instead of 3.0m above the ground, like the problem stated), then v=13.25m/s, which is equivalent to the original y component of velocity.

    Either both you and I have made the same mistake somewhere, or the website that you got the problem from has the wrong answer. Would you mind posting the answer that the website has?
  5. Sep 15, 2012 #4
    I got 15.5m/s.
  6. Sep 15, 2012 #5
    The answer 15.5m/s was actually right. Was my method of doing this
    problem was wrong or I made some mistake during calculation?
  7. Sep 15, 2012 #6
    Your method was determining the vertical velocity, which is correct.
    But at level 3m, there is also horizontal velocity.

    The question asked is the speed.
    Doing kinematic for this problem is bit messy.

    Doing with energy conservation is a bit easier since it is not a vector, just scalar.

    So the initial energy KE1=KE2+mg
    1/2m(v1)2=1/2m(v2)2 + mgh


    Actually in above equations, v is speed not velocity since energy is scalar.

    Maybe you are familiar with the last equation.
    Last edited: Sep 15, 2012

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