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Projectile motion baseball problem

  1. Sep 18, 2003 #1
    [SOLVED] projectile motion problem

    Hello guys,

    This particular problem has been giving me a headache. I can't see how to figure out this problem w/o being given an inital angle. Here it is:

    A baseball is hit at ground level. The ball reaches its max height above ground level 3 s after being hit. Then 2.5 s after reaching its max height, the ball barely clears a fence that is 97.5 m from where it was hit. Assume the ground is level. a) What max height above ground level is reached by the ball? b) How high is the fence? c) How far beyond the fence does the ball strike the ground?

    Ok, to start out I drew the picture and labeled it and everything. Then..

    Y-Yo = Vt - 1/2 at^2
    max height = m = 0(3) - 1/2 (-9.8)(3)^2 = 44.1 m

    Ok, this is where I get stuck. I'm thinking about using the trajectory equation, to get the height of the fence.

    y = (tan@)x - (gx^2)/(2(Vocos@))^2

    I'm thinking that y = the height of the fence and x is the distance to the fence. So, x = 97.5 m.

    But I don't know how to get the angle, @!

    It's been driving me nuts because I tried different ways, but kept getting answers that didn't match.

    Here's some of what I tried:

    X = Xo + Vt - 1/2 at^2
    X = 0 - 1/2 (-9.8)(3)^2 = 44.1m

    Then since X = 44.1= Y

    I did (44.1^2 + 44.1^2)^(1/2) = 62.4 m

    Then sin@ = 44.1/62.4 --> @ = about 45 degrees

    Is that correct?

    How should I approach the rest of the problem?

    Thanks much!
  2. jcsd
  3. Sep 19, 2003 #2
    Try to find the horizontal velocity and vertical velocity first before you try anything else.

    Then you can use the elapsed time to reach the maximum height and the horizontal velocity to find the total distance travelled.

    NOTE: The initial velocity is NOT zero.
  4. Sep 22, 2003 #3
    Ahhh...haven't done these problems in a while. *Feels nostalgic*

    Here's how I did your problem:

    a) Max height of ball.

    Acceleration due to gravity = -9.81ms-2
    Initial vertical velocity = unknown
    End vertical velocity = 0
    Time taken to reach end velocity = 3s

    Using v = u + at,
    0 = u + (-9.81)*3
    u = 29.43ms-1

    Then, using v^2 = u^2 + 2as
    s = 44.145m (as you got)

    b) Height of fence.

    Note: Problem is symmetrical.
    Therefore height of ball at t = 0.5s after hit is same as 5.5s after hit.

    s = unknown
    u = 29.43ms-1
    t = 0.5s
    a = -9.81ms-2

    Using s = ut + 0.5at^2
    S = 13.48875m

    c) Distance beyond fence.

    Note again: Problem is symmetrical.

    So distance = 0.5*17.72 = 8.86m
  5. Dec 1, 2009 #4
    Re: [SOLVED] projectile motion problem

    Hello, and sorry for upping an old thread.
    Could someone please tell me where the 17.72 is coming from in the above calculation ?
    I've spent the past hour trying to figure it out but I couldn't find an answer.

    Thank you very much in advance :smile:
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