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23 August, 15:14

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h (t) = - 16t2 + 40ft + 1.5. • About how many second after launch would the ball hit the ground? • What is the maximum height of the ball?

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  1. 23 August, 15:21
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    For this case we have the following equation:

    h (t) = - 16t2 + 40t + 1.5

    To hit the ground we have:

    -16t2 + 40t + 1.5 = 0

    Solving the polynomial we have:

    t1 = - 0.04

    t2 = 2.54

    Taking the positive root (for being time) we have:

    t = 2.54 s

    For the maximum height we have:

    We derive the equation:

    h ' (t) = - 32t + 40

    We equal zero and clear t:

    t = 40/32

    t = 1.25

    We evaluated t = 1.25 in the function:

    h (1.25) = - 16 * (1.25) ^ 2 + 40 * (1.25) + 1.5

    h (1.25) = 26.5 feet

    Answer:

    the ball hit the ground at:

    t = 2.54 s

    The maximum height of the ball is:

    h (1.25) = 26.5 feet
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