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 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.)

Step 1 ：We are given the height of the ball as a function of time, \(s = -2.7t^2 + 40t + 6.5\), where \(s\) is the height in feet and \(t\) is the time in seconds. We are asked to find the time \(t\) when the height \(s\) is 20 feet.

Step 2 ：This means we need to solve the equation \(-2.7t^2 + 40t + 6.5 = 20\) for \(t\). This is a quadratic equation, and we can solve it using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -2.7\), \(b = 40\), and \(c = 6.5 - 20 = -13.5\).

Step 3 ：Substituting the values of \(a\), \(b\), and \(c\) into the quadratic formula, we get two solutions for \(t\), approximately 0.35 and 14.47.

Step 4 ：These are the two times when the ball is 20 feet above the moon's surface. However, the problem asks for the time after the ball was thrown, so we should only consider the positive solution.

Step 5 ：Therefore, the ball is 20 feet above the moon's surface after approximately 0.35 seconds.

Step 6 ：Final Answer: \(\boxed{0.35}\) seconds.