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 40 ft per sec. the height s of the ball in feet is given by the equation s = - 2.7t² + 40t + 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 moons surface?
after
seconds the ball will be 10 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

Set $s = 10$ in the equation $s=-2.7t^{2}+40t + 6.5$. So we get $10=-2.7t^{2}+40t + 6.5$.

Step2: Rearrange to standard quadratic - form

Rearrange the equation to $2.7t^{2}-40t + 3.5 = 0$. Here, $a = 2.7$, $b=-40$, and $c = 3.5$.

Step3: Use the quadratic formula

The quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute the values of $a$, $b$, and $c$:
\[

$$\begin{align*} t&=\frac{40\pm\sqrt{(-40)^{2}-4\times2.7\times3.5}}{2\times2.7}\\ &=\frac{40\pm\sqrt{1600 - 37.8}}{5.4}\\ &=\frac{40\pm\sqrt{1562.2}}{5.4}\\ &=\frac{40\pm39.525}{5.4} \end{align*}$$

\]

Step4: Calculate the two values of $t$

For the plus - sign: $t_1=\frac{40 + 39.525}{5.4}=\frac{79.525}{5.4}\approx14.73$.
For the minus - sign: $t_2=\frac{40-39.525}{5.4}=\frac{0.475}{5.4}\approx0.09$.

Answer:

$0.09,14.73$