The equation h = -16t^2 + 32t + 9 gives the height of a ball, h, in feet above the ground, at t seconds after the ball is thrown upward. How many seconds after the ball is thrown will it reach its maximum height? What is its maximum height?

To find the time it takes for the ball to reach its maximum height, we can use the fact that the maximum height occurs when the ball reaches its peak, and its velocity is momentarily zero.

To do this, we need to find the derivative of the equation h = -16t^2 + 32t + 9 with respect to time (t), which will give us the velocity function. We can then set this velocity function equal to zero and solve for t.

Taking the derivative of the equation:
h = -16t^2 + 32t + 9
dh/dt = -32t + 32

Setting the velocity function equal to zero:
-32t + 32 = 0

Solving for t:
-32t = -32
t = 1

Therefore, the ball will reach its maximum height 1 second after it is thrown.

To find the maximum height, we can substitute this value of time (t=1) back into the original equation for h:
h = -16t^2 + 32t + 9
h = -16(1)^2 + 32(1) + 9
h = -16 + 32 + 9
h = 25

Therefore, the maximum height of the ball is 25 feet.

find the vertex of the parabola.

The t of the vertex tells you when it reached the maximum height, the h tells you what that height is.