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5 February, 17:23

A rock is thrown from the top of a tall building. The distance in feet between the rock and the ground t seconds after thrown is given by d = - 16t2 - 4t + 412. How long after the rock is thrown is it 407 feet from the ground?

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  1. 5 February, 18:35
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    0.448 seconds

    Step-by-step explanation:

    d = - 16t² - 4t + 412

    find t when d = 407

    substituting d = 407 into the equation:

    407 = - 16t² - 4t + 412 (subtract 407 from both sides)

    -16t² - 4t + 412 - 407 = 0

    -16t² - 4t + 5 = 0 (multiply both sides by - 1)

    16t² + 4t - 5 = 0

    solving using your method of choice (i. e completing the square or using the quadratic equation), you will end up with

    t = (-1/8) (1 + √21) = - 0.70 seconds (not possible because time cannot be negative)

    or

    t = (-1/8) (1 - √21) = 0.448 seconds (answer)
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