If a projectile is fired straight upward from the ground with an inital speed of 96 feet per second, then its height (h) in feet after t seconds is given by the equation

h(t)=-16t^2+96t

Find the maximum height of the projectile.

Tr = (V-Vo)/g.

Tr = (0-960 / -32 = 3 s. = Rise time.

h(t) = -16*3^3 + 96*3 = 144 Ft.

To find the maximum height of the projectile, we need to determine the vertex of the equation h(t) = -16t^2 + 96t.

The vertex of a quadratic equation in the form of h(t) = at^2 + bt + c can be found using the formula:

t = -b / (2a)

In this case, a = -16 and b = 96, so we substitute these values into the formula:

t = -96 / (2*(-16))
t = -96 / (-32)
t = 3

So, the projectile reaches its maximum height after 3 seconds.

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

h(t) = -16t^2 + 96t
h(3) = -16(3)^2 + 96(3)
h(3) = -16(9) + 96(3)
h(3) = -144 + 288
h(3) = 144

Therefore, the maximum height of the projectile is 144 feet.

To find the maximum height of the projectile, we need to determine the value of t that corresponds to the vertex of the quadratic equation given.

The equation h(t) = -16t^2 + 96t represents the height of the projectile at any given time t.

To find the vertex of a quadratic equation in the form ax^2 + bx + c, we use the formula t = -b / (2a). In this case, a = -16 and b = 96.

Plugging in these values into the formula, we have:

t = -96 / (2(-16))
t = -96 / (-32)
t = 3

So, the maximum height is reached at t = 3 seconds.

To find the maximum height, we substitute this value of t into the equation h(t) = -16t^2 + 96t:

h(3) = -16(3)^2 + 96(3)
h(3) = -16(9) + 288
h(3) = -144 + 288
h(3) = 144

Therefore, the maximum height of the projectile is 144 feet.