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14 August, 08:52

The height of an arrow shot upward can be given by the formula s = vnt - 16, where vo is the initial

velocity and t is time. How long does it take for the arrow to reach a height of 48 ft if it has an initial

velocity of 96 ft/s? Round to the nearest hundredth.

The equation that represents the problem is 48 = 96t - 1612.

Solve 16t - 960 + 48 = 0.

Complete the square to write 161-961 + 48 = 0 as

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Answers (1)
  1. 14 August, 09:15
    0
    I'm going to take a guess that the asker meant

    s = v₀t - 16t²

    v₀=96 ft/sec

    s=48 ft

    48 = 96t - 16t²

    16t² - 96t + 48 = 0

    Cancel a factor of 16.

    t² - 6t + 3 = 0

    Completing the square,

    (t - 3) ² - 9 + 3 = 0

    (t - 3) ² = 6

    t - 3 = ±√6

    t = 3±√6

    Round? I hate ruining a nice exact answer. Both ts are positive, one on the way up, one one the way down.

    t≈0.55 sec or t≈5.45 sec

    How long sounds like we're asking how long is the going up part, which is

    Answer: t=0.55 sec
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