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 14.68,0.14 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.)

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

Step3: Apply 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:

\(14.98\)