a) To draw the illustration, we can create a graph with the vertical distance (s) on the y-axis and time (t) on the x-axis. The equation given is s = 80t - 16t^2.
b) To determine when the ball is more than 96 feet above the ground, we can set the equation s > 96 and solve for t.
80t - 16t^2 > 96
Rearranging, we get -16t^2 + 80t - 96 > 0
Factoring out -16, we have -16(t^2 - 5t + 6) > 0
Factoring further, we get -16(t - 3)(t - 2) > 0
Since the coefficient is negative, the inequality holds true when t lies between the roots of the equation. The roots are t = 2 and t = 3.
Therefore, the ball is more than 96 feet above the ground for the time interval (2, 3).
c) To find the maximum height of the ball, we need to find the vertex of the quadratic equation s = -16t^2 + 80t.
The vertex can be found using the formula t = -b/2a. In this case, a=-16 and b=80.
t = -80 / (2*(-16))
t = -80 / (-32)
t = 2.5
The maximum height occurs at t = 2.5. To find the height, we plug this value back into the equation:
s = -16(2.5)^2 + 80(2.5)
s = -16(6.25) + 200
s = -100 + 200
s = 100
Therefore, the maximum height of the ball is 100 feet.
d) The ball reaches the maximum height after 2.5 seconds.