let t = time (in seconds); h(t) describes the height of a ball thatis thrown (in

Question

let t = time (in seconds); h(t) describes the height of a ball thatis thrown (in feet)

h(t) = -t2/12 + 2t + 4

a) initial height = 4 ft

b) max height occurs at time = -b/2a = 12 seconds
h(12) = 16 ft

c) How far from the initial position does the ball strike theground? (max range/horizontal distance the ball travels)

Can part c be solved using only precalculus w/ophysics or calculus? (i.e. w/o knowledge of velocity, derivatives,or any specialized formulas; using only the info from the previousparts on height)

Explanation / Answer

The height of the ball thrown in feet is given by              h(t) = -t2/12 + 2t + 4 (a) At the time t = 0, the height thrown is h = 0 + 0 + 4ft                                                            = 4 ft (b) At the maximum height. the value of dh / dt = 0 Then dh / dt = (-2t / 12) + 2 + 0 = 0                  (-t / 6) + 2 = 0                   t / 6 = 2                    t = 12s So at t = 12s the object reaches the maximum height. So maximum height is            hmax = [-(12)2 / 12] + 2(12) + 4                   = -12 + 24 + 4                   = 28 - 12                   = 16 ft (c) Now the maximum distance traveled is             R = 4hmax                = 4(16ft)                = 64 ft

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