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Ted Hits Homer

  1. Apr 8, 2005 #1
    Ted Williams, Professor Emeritus of Hitting with the Boston Red Sox, visits the Skydome in Toronto where all distances are measured in meters. He demonstrates to the younger Bluejays how it is done by hitting a home run that clears a low fence 120 m from home plate.

    a) What is the minimum speed with which the ball must leave the bat for the batter to have hit a home run? Neglect air resistance (which is actually not completely negligible) and assume the top of the fence is at the same height as the bat.
    vmin = m/s

    HELP: You may want to work part (b) first.

    b) At what angle to the horizontal must the ball leave the bat for this minimum speed to be sufficient? Please enter your answer in degrees.
    q = °

    c) When the ball is hit with the speed and angle determined above, how long does it take the ball to travel the distance to the fence?
    t = s

    d) For these same conditions, find the maximum height (relative to the top of the fence) reached by the ball on its trajectory.
    ymax = m

    Can you help me with some initial guidance on all these different parts..

    Thanks in advance ,
  2. jcsd
  3. Apr 8, 2005 #2
    well you should know to MAXIMIZE range your projectile (thats the kind of motion it is here) the angle of projection should be 45 degrees

    X direction
    There is no external force acting on the ball in the X direction since the ball goes horizontally, and air resistance is not included. SO the initial and final velocities must be the same, as teh ball covered teh 120m HORIZONTAL distance to the fence
    v1=v2=v0 cos 45
    d = 120
    d = vt = 120 = (v0 cos 45) t..........(1)

    The ball was initially thrown upward with a component pointing straight up. When the ball reaches the fence it has now COMPLETELY reversed direction. If i took the upward to be positive then the down must be negative Hence

    Y direction
    v1=v0 sin45
    v2=-v0 sin45
    a= -9.8m/s^2
    v2 = v1 + at
    v0 sin45 = -v0 sin45 - 9.8t
    t = 2v0 sin45 / 9.8 ......... (2)

    substitute 2 in to 1
    120 = (v0 cos 45) (2v0 sin 45 / 9.8)
    and solve for v0

    to find the time taken, well you know the velocity now, so sub it into 2 or 1 and solve for t

    To find the maximum height ask yourself what is the velocity VERTICAL ONLY when the ball has reached its maximum height?
    you know v1, v2, a (well something happened to the speed so there must a force acting) and solve for t using
    v2 = v1 + at
  4. Apr 9, 2005 #3
    At the max. height v2 = 0

    so, t = -v1/-9.81
    This gives us the time 't'


    y -y0 = v0sintheta^2 - 1/2 g *t ^2

    Is this right, vo = 34.31 m/s2 from part a.......
  5. Apr 11, 2005 #4
    Guys no luck to the last part ( Find max. height ) of this problem, please help!!! My previous post is what I did.
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