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?

I hope it was right

v = dh/dt = 32 -32 t

at top, v = 0
t = 1
h = 9+16 = 25
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now with algebra
16 t^2 -32 t =9 - h
complete square
t^2 -2 t = (-1/16)(h - 9)

t^2-2t+1 = (-1/16)(h - 9)+16/16
(t-1)^2 = -(1/16)( h-25)
t = 1, h=25 same answer

To find the maximum height of the ball, we need to determine the value of t when the height is at its maximum.

The equation h = -16t^2 + 32t + 9 represents the height of the ball. This is a quadratic equation in the form of h = at^2 + bt + c.

The coefficient of t^2 is -16, which is negative. This means that the parabolic graph opens downwards, indicating that the vertex represents the maximum point.

The formula for finding the x-coordinate of the vertex of a quadratic equation in the form h = at^2 + bt + c is given by t = -b / (2a).

In this case, a = -16 and b = 32. Substituting these values into the formula, we get:

t = -(32) / (2(-16))
t = -32 / -32
t = 1

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

To find the maximum height, we substitute t = 1 into the equation 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.

To find the time when the ball reaches its maximum height, we need to find the vertex of the equation. The vertex of a quadratic equation in the form h = -16t^2 + 32t + 9 is given by the formula:

t = -b / (2a)

In this case, a = -16 and b = 32, so substitute these values into the formula:

t = -32 / (2 * -16)

Simplifying, we have:

t = -32 / -32

t = 1

So, the ball reaches its maximum height after 1 second.

To find the maximum height, substitute this value back into the original equation:

h = -16(1)^2 + 32(1) + 9

Simplifying:

h = -16 + 32 + 9

h = 25

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