Question 332461: I have asked this ? earlier with no response.
A ball is thrown from the top of a roof 320 feet above the ground. The formula h = –16t^2 + 32t + 320 describes the height of the ball above the ground, h, in feet, t seconds after the fall begins. Find the maximum height of the ball, as measured from the ground. Hint: The path of the ball is a parabola, so think about the vertex.
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! A ball is thrown from the top of a roof 320 feet above the ground. The formula h = –16t^2 + 32t + 320 describes the height of the ball above the ground, h, in feet, t seconds after the fall begins. Find the maximum height of the ball, as measured from the ground. Hint: The path of the ball is a parabola, so think about the vertex.
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h = –16t^2 + 32t + 320
The vertex is a point (t,h) where the parabola is at its peak.
"axis of symmetry" provides the "t" value of the vertex.
The value is:
t = -b/(2a) = -32/(2(-16))
t = -32/-32
t = 1 second
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h value is found by plugging it back into the equation:
h = –16t^2 + 32t + 320
h = –16(1)^2 + 32(1) + 320
h = –16 + 32 + 320
h = 16 + 320
h = 336 feet (this is your answer)
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