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Golf ball

  1. Oct 21, 2004 #1
    Hi all - I've been trying to figure out this problem all night and I can't seem to get it right, Can someone please help me?? THANK YOU!!

    A 47.0 g golf ball is driven from the tee with an initial speed of 54.0 m/s and rises to a height of 23.8 m.

    (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.

    (b) What is its speed when it is 10.0 m below its highest point?
  2. jcsd
  3. Oct 21, 2004 #2
    conserve energy:

    (A) at its heighest point the potential energy is simply mgh. it will loose all the kinetic energy.

    47*9.8*23.8 and the kinetic energy is zero.

    (B) at 10 m below its highest point its speed will be.

    K(initial)=K(at 10m)+U(at 10m), U is the potential energy, and K is the kinetic.


    1/2mv^2 = 1/2mV^2 + mgh.

    lets make life easy and get rid of m from both sides.


    1/2 54^2 = 1/2 V^2 + 9.8*(23.8-10m)

    solve for V, and you will have your anwser.
  4. Oct 22, 2004 #3
    But what about the horizontal velocity?

    But part A asks:

    "Neglect air resistance and determine the kinetic energy of the ball at its highest point."

    "cyrusabdollahi's" post is valid for vertical kinetic energy at the highest point. But what about the horizontal kinetic energy? The total velocity of the ball is reduced as a result of the loss of speed from gravitational forces applied through to the top of the arc but the horizontal velocity component is still there and has its own kinetic energy.

    To detemine the vertical component of the ball's velocity off the tee, we can use

    V^2 = v^2 + 2as

    Big V in this case is the velocity at the top of the arc. Little v is the vertical component of the velocity at impact, a is g, and s is the height travelled.

    Therefore at the top V^2 is zero, leaving

    -(v^2) = 2as or -v = sqrt(2*9.81*23.8) = -21.6 m/s (relative to the gravitational component)

    Using the Pythagorean theorem to resolve into vertical and horizontal components you get 49.5 m/s for the horizontal speed at impact and, therefore, in the absence of wind resistance, the horizontal speed (and only speed component) at the top of the arc.

    Then the horizontal KE = (1/2)mv^2 = 1/2(0.047kg)(49.5m/s)^2 = 57.6 J

    On to part B

    That question also does not distinguish speed as being expressly vertical, either.

    The vertical speed at 10m below the highest point is the vertical speed at a height of 23.8-10 = 13.8m


    V^2 = v^2 +2as = -(21.6)^2 m/s + 2(9.8)(13.8) = -13.98 m/s (relative to earth's gravity) where big V is the vertical speed 10 meters below the highest point, and little v is the initial vertical speed at impact, and, again, a is gravity and s is the height.

    Finally using the pythagorean theorem again, total velocity at 10 meters below is = sqrt(13.98^2 + 49.49^2) = 51.4 m/s
    Last edited: Oct 22, 2004
  5. Dec 2, 2011 #4
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