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30 April, 21:35

Recall that the height of an object t seconds after it begins moving when gravity is the only force acting on the object is given by

h (t) = - 16t2+v0t+h0,

where h (t) is given in feet, v0 is the initial velocity of the object in ft/sec, and h0

is the initial height of the object in feet.

If a ball is thrown directly upward from the top of a building with an initial velocity of 60

feet per second, and an initial height of 75 feet. What is the maximum height attained by the ball?

Your answer is feet.

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  1. 30 April, 21:50
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    Y = - 16 t^2 + Vo t + Ho

    Vo = 60 feet/s

    Ho = 75 feet

    y = - 16 t^2 + 60 t + 75

    The maximum heigth correspond to the vertex of the parabola

    You can find it by finding the roots using the quadratic equation (which I let to you) or by finding the highest value using derivatives (which I do next):

    y' = - 32 t + 60 = 0 = > t = 60/32 = 1.875

    y = - 16 (1.875) ^2 + 60 (1.875) + 75 = 131.25 feet.

    Answer: 131.25 feet.
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