Question

watch the video and then solve the problem given below click here to watch the video 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 20 ft above the moons surface? after ( square ) seconds the ball will be 20 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 \(s = 20\) in the equation \(s=-2.7t^{2}+40t + 6.5\). So we get \(20=-2.7t^{2}+40t + 6.5\).
Rearrange it to the standard quadratic form \(ax^{2}+bx + c=0\).
\(2.7t^{2}-40t + 13.5 = 0\), where \(a = 2.7\), \(b=-40\), \(c = 13.5\).

Step2: Use the quadratic formula

The quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute \(a = 2.7\), \(b=-40\), \(c = 13.5\) into the formula:
\(t=\frac{40\pm\sqrt{(-40)^{2}-4\times2.7\times13.5}}{2\times2.7}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-40)^{2}-4\times2.7\times13.5=1600 - 145.8=1454.2\).
Then \(t=\frac{40\pm\sqrt{1454.2}}{5.4}\).
\(\sqrt{1454.2}\approx38.13\).
\(t_{1}=\frac{40 + 38.13}{5.4}=\frac{78.13}{5.4}\approx14.47\).
\(t_{2}=\frac{40-38.13}{5.4}=\frac{1.87}{5.4}\approx0.35\).

Answer:

\(0.35,14.47\)