Question 268526
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I'm reasonably certain that you meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ 224t\ -\ 16t^2]


That is because, in general, the height of an object propelled upward from a start position of *[tex \Large h_o] at an initial velocity magnitude of *[tex \Large v_o] as a function of time is given by:


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


Here, you are given that *[tex \Large h_o\ =\ 0] and *[tex \Large v_o\ =\ 224], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ -16t^2\ +\ 224t]


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h(t)\ =\ 224t\ -\ 16t^2]


This is a quadratic equation having two roots.  Clearly, since the equation has no constant term (i.e. *[tex \Large h_o\ =\ 0]), one of the roots will be zero.  That makes good sense since the object began at height zero at time zero.  Now all you need to do is solve for the other root to find the other time that the height will be zero, namely the time in the future when the projectile comes back to the ground.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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