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 t 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 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 14 47,0.35 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.) b. how many seconds will it take for the ball to hit the moons surface? it will take ( square ) seconds for the ball to hit the moons surface (round to the nearest hundredth as needed.)

Explanation:

Step1: Set up the equation for part b

When the ball hits the moon's surface, \(s = 0\). So we set the equation \(-2.7t^{2}+40t + 6.5=0\). For a quadratic equation \(ax^{2}+bx + c = 0\) (here \(a=-2.7\), \(b = 40\), \(c = 6.5\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step2: Calculate the discriminant

First, calculate the discriminant \(\Delta=b^{2}-4ac=(40)^{2}-4\times(-2.7)\times6.5=1600 + 70.2=1670.2\).

Step3: Substitute into the quadratic formula

\(t=\frac{-40\pm\sqrt{1670.2}}{2\times(-2.7)}=\frac{-40\pm40.87}{-5.4}\).
We have two solutions for \(t\):

• \(t_1=\frac{-40 + 40.87}{-5.4}=\frac{0.87}{-5.4}\approx - 0.16\) (rejected since time \(t\geq0\)).
• \(t_2=\frac{-40-40.87}{-5.4}=\frac{-80.87}{-5.4}\approx14.98\).

Answer:

\(14.98\)