Step 1 :We are given the height equation of the ball as \(s = -2.7t^2 + 40t + 6.5\), where \(s\) is the height in feet and \(t\) is the time in seconds.
Step 2 :For part a, we need to find the time when the ball is 20 feet above the moon's surface. We set \(s\) to 20 and solve for \(t\).
Step 3 :Solving the equation \(-2.7t^2 + 40t + 6.5 = 20\) gives us two solutions for \(t\), approximately 0.35 seconds and 14.47 seconds.
Step 4 :So, the ball is 20 feet above the moon's surface at approximately \(\boxed{0.35, 14.47}\) seconds.
Step 5 :For part b, we need to find the time when the ball hits the moon's surface. This is when \(s\) is 0. So, we set \(s\) to 0 and solve for \(t\).
Step 6 :Solving the equation \(-2.7t^2 + 40t + 6.5 = 0\) gives us two solutions for \(t\), approximately -0.16 seconds and 14.98 seconds.
Step 7 :However, time cannot be negative, so we discard the negative solution. Therefore, the ball hits the moon's surface at approximately \(\boxed{14.98}\) seconds.