SOLUTION: 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

Question 215961This question is from textbook
: 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.
a) For what t>0 is the object feet from the ground?

b)For what t.0 is the object on the ground? This question is from textbook

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
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?
-----------------------
b)For what t.0 is the object on the ground?
-16t^2 + 64t + 650 = 0
-8t^2 + 32t + 325 = 0

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=11424 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -4.68019460794369, 8.68019460794369. Here's your graph:

-----------
t = ~ 8.68 seconds

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