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A ball is thrown vertically upward from ground level with initial velocity of 96 feet per second. Assume the acceleration of the ball is a(t) = -32 ft^2 per second. (Neglect air Resistance.)

(a) How long will it take the ball to raise to its maximum height? What is the maximum heights?

(b) After how many seconds is the velocity of the ball one-half the initial velocity?

(c) What is the height of the ball when its velocity is one-half the initial velocity?

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2 answers

  1. (a) Acceleration is Change in velocity over time, thus we can write it as,
    a = (vf - vo) / t
    Note that final velocity (vf) is Zero when it reaches its Maximum height since it stops. And since acceleration and Initial velocity (vo) are given, we can Substitute:
    -32 = (0 - 96) / t
    -32t = -96
    t = 3 seconds

    For Maximum height, we can use the formula,
    h = vo*t - (1/2)(a)(t^2)
    h = 96*3 - (1/2)(-32)(3^2)
    h = 144 ft

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  2. (c) We can calculate for the height of the ball when it is 1/2 of Initial velocity using the formula,
    vf^2 - vo^2 = 2gd
    Since vf= 1/2*vo,
    (1/2*vo)^2 - vo^2 = 2gd
    Substituting,
    (1/2*96)^2 - 96^2 = 2*(-32)*d
    48^2 - 96^2 = -64d
    d = -6912/-64
    d = 108 ft

    (b) We can calculate the time when the ball is moving at 1/2 of its Initial velocity with the same formula we used in (a),
    h = vo*t - (1/2)(a)(t^2)
    Substitute h = 108, and solve for t.

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