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 14 ft above the moons surface?

after ( square ) seconds the ball will be 14 ft above the moons surface.

(round to the nearest hundredth as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Substitute \(s = 14\) into the equation

Substitute \(s = 14\) into \(s=-2.7t^{2}+40t + 6.5\), we get \(14=-2.7t^{2}+40t + 6.5\).
Rearrange it to the standard quadratic form \(ax^{2}+bx + c = 0\): \(2.7t^{2}-40t+7.5 = 0\). Here \(a = 2.7\), \(b=-40\), \(c = 7.5\).

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

First, calculate the discriminant \(\Delta=b^{2}-4ac=(-40)^{2}-4\times2.7\times7.5=1600 - 81=1519\).
Then \(t=\frac{40\pm\sqrt{1519}}{2\times2.7}\).
\(\sqrt{1519}\approx38.97\).
\(t_1=\frac{40 + 38.97}{5.4}=\frac{78.97}{5.4}\approx14.62\).
\(t_2=\frac{40-38.97}{5.4}=\frac{1.03}{5.4}\approx0.19\).

Answer:

\(0.19,14.62\)