A ball is thrown from the top of a 50-ft building with an upward velocity of 24 ft/s. When will it reach its maximum height? How far above the ground will it be?

1 answer

as with all quadratics, the vertex is at t = -b/2a
so, since our equation for height is
h(t) = -16t^2 + 24t + 50
the vertex (max height) is at t = 24/32 = 3/4
Now just find h(3/4)