Question:

Gravity and time - Kinematics?

by Guest56180  |  earlier

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A baseball is hit nearly straight up into the air with a speed of 33 m/s.

How high does it go?

How long is it in the air?

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  1. Vmax = V(0) -gt, where v(0) is the initial velocity, and V max is the final velocity which the ball reaches upwards when it's velocity becomes equal to zero.

    V 0 = g*t

    33 = 9.8*t

    t = 33/9.8 = 3.367 seconds

    The maximum height the ball may reach is after 3.367 seconds.

    Y = V(0)*t -0.5*g*(t)^2

    Y = v(0)*t -0.5*9.8*(3.367)^2

    Y = 33*3.367 - 0.5*9.8*3.367^2

    Y = 55.55 meters

    The ball stays in the air an interval =2*t=2*3.367=6.734 seconds


  2. initial v = 33 m/s

    acceleration = -g = - 9.8 m/s^2

    How Long in the Air:

    y = v*t + 1/2*a*t^2

    y = 33t - 4.9t^2

    Since initial y = final y = 0, then:

    0 = 33t - 4.9t^2

    Using the quadratic formula: t = 6.73 seconds

    How High:

    Since the ball is in the air 6.73 seconds, then it must hit its apex at 1/2t seconds, or t = 3.367 s.

    y = 33 (3.367) - 4.9 (3.367^2)

    y = 55.56 m

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