Question 977220
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Height as a function of time for a projectile with a vertical velocity component is:


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


where *[tex \Large v_o] is the initial velocity and *[tex \Large h_o] is the initial height.  Your *[tex \Large v_o] is 96 ft/sec and your *[tex \Large h_o] is 52.  Solve the quadratic for the positive root which is the value of *[tex \Large t] when the projectile hits the ground.


Since velocity is the rate of change of position, take the first derivative of the the height function and evaluate at the time the projectile hits the ground.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \