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12 August, 03:59

ted popped a baseball straight up with an initial velocity of 48 ft/s. The height, h, in feet, of the ball above the ground is modeled by the function h (t) = -16t2 + 48t+3. How long was the ball in the air if the catcher the ball 3 feet above the ground?

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  1. 12 August, 06:04
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    3 seconds

    Step-by-step explanation:

    We have to give us the following function h (t) = - 16 * t ^ 2 + 48 * t + 3

    Now, they tell us that the height is 3 feet, therefore replacing that value we have left:

    3 = - 16 * t ^ 2 + 48 * t + 3, operating cancels 3. And it would be:

    -16 * t ^ 2 + 48 * t = 0

    Removing common factor - 16 * t, it would be:

    (-16 * t) * (t - 3) = 0

    We have two solutions:

    -16 * t = 0 - --> t = 0

    t - 3 = 0 - --> t = 3

    Because the time cannot be 0 seconds, the time it would take the ball in the air is 3 seconds.
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