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1 April, 11:05

If an object is propelled upward from a height of s feet at an initial velocity of v feet per second, then its height h after t seconds is given by the equation h=-16 t squared + vt+ s, where h is in feet. If the object is propelled from a height of 12 feet with an initial velocity of 64 feet per second, its height h is given by the equation h=-16t squared+64t+12. After how many seconds is the height 72 feet?

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  1. 1 April, 11:38
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    From the given information we can write the equation:

    -16t^2 + 96t + 4 = 112

    -16t^2 + 96t + 4 - 112 = 0

    -16t^2 + 96t - 108 = 0

    simplify, change the signs, divide by - 4

    4t^2 - 24t + 27 = 0

    You can use the qudratic formula but this will factor to

    (2t-3) (2t-9) = 0

    Two solutions

    t = 3/2

    t = 1.5 seconds at 112 ft on the way up

    and

    t = 9/2

    t = 4.5 seconds at 112 ft on the way back down

    Graphically, (green line is 112 ft)
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