Question 1169278
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  h(t)\ =\ 16t^2\ +\ v_ot\ +\ h_o]


Where *[tex \Large v_o] is the given initial velocity and *[tex \Large h_o] is the height of the building plus the distance the person's hand holding the ball was above the top of the building when the ball was released


The ball never hits the ground.  The problem is posed as the ball is thrown vertically upward and no force with a horizontal component is mentioned in the problem statement.  Therefore the ball goes straight up and comes straight back down again impacting the roof of the building.


Given that the ball impacts the roof of the building, adjust the initial height to be the height above the roof that the ball is released, set the function equal to zero, and solve for the positive root to find the time that the ball impacts the roof on the way down.  Then, to find the impact velocity evaluate the first derivative of the function at the time of impact.


Calculate the height ignoring the height of the building, then the distance traveled is twice the height minus the height of the hand above the roof when the ball is released.


I'm sure this is not the answer your instructor expected, but it is 100% correct based on the way the question is worded.

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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