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26 November, 20:54

Two stones are thrown vertically upward with matching initial velocities of 48ft/s at timet = 0. One stone is thrown from the edge of a bridge that is 32 ftabove the ground and the other stone is thrown from ground level. The heightof the stone thrown from the bridge aftertseconds isf (t) = -16t2 + 48t + 32, and the height of the stone thrown from the ground aftertseconds isg (t) = -16t2 + 48t.

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  1. 26 November, 21:16
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    Step-by-step explanation:

    Stone thrown from ground:

    S = - 16t² + 48t + 32

    Velocity = V = dS/dt = - 32t + 48

    Acceleration = dV/dr = - 32

    Because Velocity is 0 at the top, we have V = 0

    That is

    -32t + 48 = 0

    t = 48/32 = 3/2

    S = - 16 (3/2) ² + 48 (3/2) + 32

    = 36 + 72 + 32

    S = 140 ft ... (1)

    Stone thrown above the bridge level:

    S = - 16t² + 48t

    Velocity = V = dS/dt = - 32t + 48

    Acceleration = dV/dr = - 32

    Because Velocity is 0 at the top, we have V = 0

    That is

    -32t + 48 = 0

    t = 48/32 = 3/2

    S = - 16 (3/2) ² + 48 (3/2)

    = 36 + 72

    S = 108ft ... (2)

    From (1) and (2)

    140 - 108 = 32ft

    Which shows that the stone then from the ground levels is 32ft higher than the one thrown at the edge of the bridge.
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