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