SOLUTION: An object is projected vertically upward from the top of a building with an initial velocity of 96 ft/sec. Its distance s(t) in feet above the ground after t seconds is given by th

Question 840522: An object is projected vertically upward from the top of a building with an initial velocity of 96 ft/sec. Its distance s(t) in feet above the ground after t seconds is given by the equation
s(t) = −16t2 + 96t + 120.
(a) Find its maximum distance above the ground.
s(t) = ft
(b) Find the height of the building.
Found 2 solutions by josmiceli, amalm06:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
The formula for the vertex ( or peak ) is

Now plug the value of back
into the equation to find

(a)
The max distance above ground is 264 ft
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To find the height of the building, set

(b)
the height of the building is 120 ft
---------------------------------
Here's the plot:

Answer by amalm06(224) (Show Source): You can put this solution on YOUR website!
The velocity of the object is given by

The maximum distance above the ground occurs when the velocity of the ball is 0.

-32t+96=0, therefore t=3. The object reaches its maximum height after 3 s.

a) (Answer)

For b), the ball is at the top of the building at time t=0.

(Answer)
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