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 30 ft per sec . The height s of the ball in feet is given by the equation \( s=-2.7 \mathrm{t}^{2}+30 \mathrm{t}+6.5 \), 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 10 ft above the moon's surface?

To find the time when the ball is 10 ft above the moon's surface, we need to set the height equation equal to 10 and solve for \( t \). Given equation: \( s = -2.7t^2 + 30t + 6.5 \) We want to find when the height \( s \) is 10 ft, so we set \( s = 10 \) and solve for \( t \): \( 10 = -2.7t^2 + 30t + 6.5 \) Now, we can solve this quadratic equation to find the value of \( t \) when the ball is 10 ft above the moon's surface. Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(10=-2.7t^{2}+30t+6.5\) - step1: Convert the decimal into a fraction: \(10=-\frac{27}{10}t^{2}+30t+\frac{13}{2}\) - step2: Swap the sides: \(-\frac{27}{10}t^{2}+30t+\frac{13}{2}=10\) - step3: Move the expression to the left side: \(-\frac{27}{10}t^{2}+30t-\frac{7}{2}=0\) - step4: Multiply both sides: \(\frac{27}{10}t^{2}-30t+\frac{7}{2}=0\) - step5: Multiply both sides: \(10\left(\frac{27}{10}t^{2}-30t+\frac{7}{2}\right)=10\times 0\) - step6: Calculate: \(27t^{2}-300t+35=0\) - step7: Solve using the quadratic formula: \(t=\frac{300\pm \sqrt{\left(-300\right)^{2}-4\times 27\times 35}}{2\times 27}\) - step8: Simplify the expression: \(t=\frac{300\pm \sqrt{\left(-300\right)^{2}-4\times 27\times 35}}{54}\) - step9: Simplify the expression: \(t=\frac{300\pm \sqrt{300^{2}-3780}}{54}\) - step10: Simplify the expression: \(t=\frac{300\pm 6\sqrt{2395}}{54}\) - step11: Separate into possible cases: \(\begin{align}&t=\frac{300+6\sqrt{2395}}{54}\\&t=\frac{300-6\sqrt{2395}}{54}\end{align}\) - step12: Simplify the expression: \(\begin{align}&t=\frac{50+\sqrt{2395}}{9}\\&t=\frac{300-6\sqrt{2395}}{54}\end{align}\) - step13: Simplify the expression: \(\begin{align}&t=\frac{50+\sqrt{2395}}{9}\\&t=\frac{50-\sqrt{2395}}{9}\end{align}\) - step14: Rewrite: \(t_{1}=\frac{50-\sqrt{2395}}{9},t_{2}=\frac{50+\sqrt{2395}}{9}\) The solutions to the quadratic equation are \( t_1 = \frac{50 - \sqrt{2395}}{9} \) and \( t_2 = \frac{50 + \sqrt{2395}}{9} \). Therefore, the ball is 10 ft above the moon's surface after \( t_1 \) and \( t_2 \) seconds.