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Problem: Given Angle and Distance

  1. Sep 13, 2007 #1
    A baseball leaves the bat at 31 degrees of the horizontal. The ball was caught by an outfielder 400 ft from the home plate. The height where the ball was hit is the same where it was caught by the outfielder.

    whats the initial speed of the ball?
    whats the maximum height it reaches?

    2. Relevant equations
    1. y = y0 + voyt + (1/2)(g)(t)^2
    2. vy = v0y + (g)t

    3. The attempt at a solution
    This is where it got a little messy.
    First off:
    I converted 400 ft to meters getting 121.9 m.
    After that, from the pythagorean theorem: See pic 1.

    After, i figured that at the half distance of 121.9m (60.95m) the Vy = 0.
    I then used the theorem of ratio of triangles to find the maximum height would be if gravity is 0. The max height would be 36.6 m. see pic 2.

    Since i don't have the time, i got t from equation 2 and substituted it in equation 1. I eventually derived (v0y)^2 (30/196) = y

    Substitute y for 36.6m
    i got v0y = 15.46m

    and from there I just derived everything else from that number.

    v0 = 30 m/s
    max height = 12.17 m

    help is appreciated. Thanks in advance.

    Attached Files:

  2. jcsd
  3. Sep 13, 2007 #2


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    Homework Helper

    The ball does not follow a straight line path along 31 degrees... so you can't use the pythogrean theorem to get the maximum height.

    What is the equation for horizontal displacement in terms of time, v0, theta...

    What is the equation for vertical displacement in terms of time, v0, theta...