A projectile is thrown upward so that its distance above the ground after t seconds is h(t)=-16t^2 + 570t. After how many seconds does it reach its maximum​ height?

To find the maximum height of the projectile, we need to find the vertex of the parabola represented by the equation h(t) = -16t^2 + 570t.

The vertex of a parabola in the form of y = ax^2 + bx + c is given by the formula:

x = -b / (2a)

In our case, a = -16 and b = 570.

Substituting the values into the formula:

t = -570 / (2 * -16)

Simplifying the equation:
t = -570 / -32
t = 17.8125

Therefore, the projectile reaches its maximum height after approximately 17.8125 seconds.

To find the maximum height, we need to find the vertex of the parabolic function h(t) = -16t^2 + 570t. The vertex of a parabola can be found using the formula t=-b/2a, where a and b are the coefficients of t^2 and t, respectively.

In this case, a = -16 and b = 570. Plugging these values into the formula, we get:

t = -570 / (2 * -16)
t = -570 / -32
t = 17.8125

This means that the projectile reaches its maximum height after approximately 17.8125 seconds.

570 feet/second ????

You threw it up at about 400 miles per hour, typo but anyway

where is the vertex of that parabola?
16 t^2 -570 t = -h

t^2 - 265/8 t +265^2/16^2 = -h+265^2/16^2

(t-265/16)^2 = etc
so vertex at t = 265/16
BUT
I do not believe you

vertex at t = -b/2a = 570/32 = 285/16