Question 215961
The height of an object is given by h=-16t^2+v^0t+h^0 ft. after t seconds, where v^0 is the initial velocity of the object in ft./sec and h^0 is the initial height of the object in feet. 
An object is thrown upwards from the top of a 650ft. building at an initial velocity of 64 ft./sec. 
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h(t) = -16t^2 + 64t + 650
a) For what t>0 is the object feet from the ground?
?? How many feet?
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b)For what t.0 is the object on the ground?
-16t^2 + 64t + 650 = 0
-8t^2 + 32t + 325 = 0
*[invoke solve_quadratic_equation -8,32,325]
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t = ~ 8.68 seconds