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Physics

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A baseball is hit at ground level. The ball is observed to reach its maximum height above ground level 3.0 s after being hit. And 2.5 s after reaching this maximum height, the ball is observed to barely clear a fence that is 97.5 m from where it was hit. How high is the fence?

  • Physics - ,

    Horizontal:
    97.5=vi*5.5
    so vihorizontal= 97.5/5.5 m/s

    vertical: at the top, the vertical veloicty is zero, so
    0=viy-1/2 g*3^3
    viy=4.9*9g

    so now we have the vertical and horizontla components.

    finally
    hf=viy^t-1/2 g t^2 so at t=5.5, knowing viy, calculate hf.

  • Physics - ,

    5.5

  • Physics - ,

    First, you want to find the maximum height.
    At the maximum height, you know the velocity must be 0.

    vi = initial velocity
    vy(t) = vertical velocity
    t1 = first time (3 s)

    SO:
    vy(t1) = -g*t1 + vi*sinθ
    0 = -g*t1 + vi*sinθ
    vi*sinθ = g*t1

    Now, it's just a matter of substituting that into the equation for vertical distance.

    ymax = maximum height reached by ball
    y(t) = vertical distance

    ymax = y(t1) = -1/2*g*t1^2 + vi*sinθ*t1
    = -1/2*g*t1^2 + g*t1*t1
    = -1/2*g*t1^2 + g*t1^2
    = 1/2*g*t1^2
    = 1/2(9.8)3^2 = 44.1m

    From there, it's a simple matter of finding vi*sinθ.

    44.1 = -1/2*g*t1^2 + vi*sinθ*t1
    44.1 = -1/2(9.8)3^2 + vi*sinθ(3)
    vi*sinθ = 29.4

    When t = 5.5
    y(5.5) = -1/2(9.8)5^2 + 29.4(5.5)
    = 13.475m

    Thus, in order for the ball to clear the fence, it must be less than 13.475 m!

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