Question

an astronaut on the moon throws a baseball upward. the astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 50 ft per sec. the height s of the ball in feet is given by the equation s = - 2.7t²+50t + 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 12 ft above the moons surface?
after □ seconds the ball will be 12 ft above the moons surface. (round to the nearest hundredth as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Set up the height - equation

The height equation of the ball is \(s=-2.7t^{2}+50t + 6.5\), and we want to find \(t\) when \(s = 12\). So we set up the quadratic equation \(-2.7t^{2}+50t+6.5 = 12\), which can be rewritten as \(-2.7t^{2}+50t - 5.5=0\). For a quadratic equation \(ax^{2}+bx + c = 0\) (here \(a=-2.7\), \(b = 50\), \(c=-5.5\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step2: Calculate the discriminant \(\Delta=b^{2}-4ac\)

\(\Delta=(50)^{2}-4\times(-2.7)\times(-5.5)=2500 - 59.4 = 2440.6\).

Step3: Calculate \(t\) using the quadratic formula

\[

$$\begin{align*} t&=\frac{-50\pm\sqrt{2440.6}}{2\times(-2.7)}\\ &=\frac{-50\pm49.4}{-5.4} \end{align*}$$

\]
First, consider the plus - sign: \(t_1=\frac{-50 + 49.4}{-5.4}=\frac{-0.6}{-5.4}\approx0.11\). Second, consider the minus - sign: \(t_2=\frac{-50-49.4}{-5.4}=\frac{-99.4}{-5.4}\approx18.41\).

Answer:

\(t\approx0.11,18.41\)