SOLUTION: A person standing close to the edge on top of a 40 foot building throws a bal vertically upward. The quadratic function h(t)= -16t^2 +72t + 40 models the ball's height about the gr
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Question 1160555: A person standing close to the edge on top of a 40 foot building throws a bal vertically upward. The quadratic function h(t)= -16t^2 +72t + 40 models the ball's height about the ground, h(t), in feet, t seconds after it was thrown.
What is the maximum height of the ball?
How many seconds does it take until the ball hits the ground?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The maximum height is at the vertex of the parabola described by the quadratic function.
The vertex of \ =\ ax^2\ +\ bx\ +\ c)
is at \))
. For your height function, the maximum height is the 
-coordinate of the vertex point.
The height of the ball when it hits the ground is zero feet. Solve:
for the positive root.
John
My calculator said it, I believe it, that settles it
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