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 18 ft above the moons surface?
after 0.23,18.23 seconds the ball will be 18 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 equation

We are given the height - equation \(s=-2.7t^{2}+50t + 6.5\) and we want to find \(t\) when \(s = 18\). So we set up the quadratic equation \(-2.7t^{2}+50t+6.5 = 18\), which can be rewritten as \(-2.7t^{2}+50t - 11.5=0\). For a quadratic equation \(ax^{2}+bx + c = 0\), here \(a=-2.7\), \(b = 50\), and \(c=-11.5\).

Step2: Use the quadratic formula

The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(50)^{2}-4\times(-2.7)\times(-11.5)=2500 - 124.2 = 2375.8\).

Step3: Calculate the values of \(t\)

\[

$$\begin{align*} t&=\frac{-50\pm\sqrt{2375.8}}{2\times(-2.7)}\\ &=\frac{-50\pm48.742}{-5.4} \end{align*}$$

\]
For the plus - sign: \(t=\frac{-50 + 48.742}{-5.4}=\frac{-1.258}{-5.4}\approx0.23\). For the minus - sign: \(t=\frac{-50 - 48.742}{-5.4}=\frac{-98.742}{-5.4}\approx18.23\).

Answer:

\(0.23,18.23\)