1) We can rewrite the function in vertex form to find the time when the ball hits the ground. To do this, we need to complete the square.
h(t) = -5t^2 + 40t
h(t) = -5(t^2 - 8t)
h(t) = -5(t^2 - 8t + 16) + 80 (adding and subtracting 16 inside the parentheses to complete the square)
h(t) = -5(t - 4)^2 + 80 (factoring and simplifying)
So, the function in vertex form is h(t) = -5(t - 4)^2 + 80.
2) To find when the ball hits the ground, we need to find when h(t) = 0, since h(t) represents the height of the ball above the ground.
0 = -5(t - 4)^2 + 80 (substituting h(t) = 0)
5(t - 4)^2 = 80
(t - 4)^2 = 16
t - 4 = ±4
t = 4 ± 4
So, the ball hits the ground at t = 0 or t = 8 seconds after launch. However, we can discard the t = 0 solution because that represents the time when the ball is launched, and we are looking for when it hits the ground. Therefore, the ball hits the ground at t = 8 seconds after launch.