Question 536600
A person standing cloes to the edge on the top of a 200-foot building throws a baseball vertically upward. The quadratic function

                        s(t)=-16t^2+64t+200

models the ball's height above the ground, s(t), in feet, t seconds after it was thrown.
A)After how many seconds does the ball reach it's [sic] maximum height? What is the maximum height?
The equation is a parabola, the max height is at the vertex.
The vertex is on the line of symmetry, which is
t = -b/2a
t = -64/-32 = 2 seconds
s(2) = -16*4 + 64*2 + 200
max ht = 264 ft
--------------------
B)How many seconds does it take until the ball finally hits the ground?
Find t when s(t) = 0
-16t^2 + 64t + 200 = 0
2t^2 - 8t - 25 = 0
*[invoke solve_quadratic_equation 2,-8,-25]
-----------
Ignore the negative solution
t =~ 6.062 seconds
----------------
PS  it's = it is