1 point(s) possible 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 \( \mathrm{s}=-2.7 \mathrm{t}^{2}+30 \mathrm{t}+6.5 \), where \( t \) is the number of seconds after the ball was thrown. Complete parts a and b .

To analyze the motion of the baseball thrown by the astronaut on the moon, we can use the given height equation: \[ s(t) = -2.7t^2 + 30t + 6.5 \] where: - \( s(t) \) is the height of the ball in feet, - \( t \) is the time in seconds after the ball was thrown. ### Part a: Find the maximum height of the ball. To find the maximum height, we need to determine the vertex of the parabola represented by the equation. The vertex \( t \) value for a quadratic equation in the form \( at^2 + bt + c \) can be found using the formula: \[ t = -\frac{b}{2a} \] In our equation, \( a = -2.7 \) and \( b = 30 \). Plugging in these values: \[ t = -\frac{30}{2 \times -2.7} = \frac{30}{5.4} \approx 5.56 \text{ seconds} \] Now, we can substitute this \( t \) value back into the height equation to find the maximum height: \[ s(5.56) = -2.7(5.56)^2 + 30(5.56) + 6.5 \] Calculating \( (5.56)^2 \): \[ (5.56)^2 \approx 30.9136 \] Now substituting back into the equation: \[ s(5.56) = -2.7(30.9136) + 30(5.56) + 6.5 \] \[ s(5.56) \approx -83.46 + 166.8 + 6.5 \approx 89.84 \text{ feet} \] Thus, the maximum height of the ball is approximately **89.84 feet**. ### Part b: Determine when the ball hits the ground. To find when the ball hits the ground, we need to set \( s(t) = 0 \) and solve for \( t \): \[ -2.7t^2 + 30t + 6.5 = 0 \] We can use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = -2.7 \), \( b = 30 \), and \( c = 6.5 \). Plugging in these values: \[ t = \frac{-30 \pm \sqrt{30^2 - 4 \cdot (-2.7) \cdot 6.5}}{2 \cdot (-2.7)} \] \[ t = \frac{-30 \pm \sqrt{900 + 70.8}}{-5.4} \] \[ t = \frac{-30 \pm \sqrt{970.8}}{-5.4} \] \[ t = \frac{-30 \pm 31.11}{-5.4} \] Calculating the two possible values for \( t \): 1. \( t = \frac{-30 + 31.11}{-5.4} \approx \frac{1.11}{-5.4} \approx -0.205 \) (not valid since time cannot be negative) 2. \( t = \frac{-30 - 31.11}{-5.4} \approx \frac{-61.11}{-5.4} \approx 11.31 \) Thus, the ball hits the ground approximately **11.31 seconds** after it was thrown.