An astronaut on the moon throws a baseball upward. The astronaut is $6 \mathrm{ft}, 6$ in. tall, and the initial velocity of the ball is $40 \mathrm{ft}$ per sec. The height s of the ball in feet is given by the equation $\mathrm{s}=-2.7 \mathrm{t}^{2}+40 \mathrm{t}+6.5$, where $t$ is the number of seconds after the ball was thrown. Complete parts a and $b$.
a. After how many seconds is the ball $20 \mathrm{ft}$ above the moon's surface?
After $0.35,14.47$ seconds the ball will be $20 \mathrm{ft}$ above the moon's surface.
(Round to the nearest hundredth as needed. Use a comma to separate answers as needed.)
b. How many seconds will it take for the ball to hit the moon's surface?
It will take seconds for the ball to hit the moon's surface. (Round to the nearest hundredth as needed.)

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.