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 = - 2.7t^{2}+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 12 ft above the moons surface?
after ( square ) 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: Substitute \(s = 12\) into the equation

We have the equation \(12=-2.7t^{2}+40t + 6.5\).
Rearrange it to the standard quadratic form \(ax^{2}+bx + c = 0\).
\(2.7t^{2}-40t+5.5 = 0\). Here \(a = 2.7\), \(b=-40\), \(c = 5.5\).

Step2: Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

First, calculate the discriminant \(\Delta=b^{2}-4ac=(-40)^{2}-4\times2.7\times5.5\)
\(=1600 - 59.4=1540.6\).
Then \(t=\frac{40\pm\sqrt{1540.6}}{2\times2.7}=\frac{40\pm39.25}{5.4}\).

Step3: Calculate the two values of \(t\)

For the plus - sign: \(t_{1}=\frac{40 + 39.25}{5.4}=\frac{79.25}{5.4}\approx14.68\).
For the minus - sign: \(t_{2}=\frac{40-39.25}{5.4}=\frac{0.75}{5.4}\approx0.14\).

Answer:

\(0.14,14.68\)