an astronaut on the moon throws a baseball upward. the astronaut is 6 ft, 6 in. tall, and the initial…

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 = - 2.7t² + 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 10 ft above the moons surface? after 0.09, 14.73 seconds the ball will be 10 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 seconds for the ball to hit the moons surface (round to the nearest hundredth as needed.)

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 = - 2.7t² + 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 10 ft above the moons surface? after 0.09, 14.73 seconds the ball will be 10 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 seconds for the ball to hit the moons surface (round to the nearest hundredth as needed.)

Answer

Explanation:

Step1: Set up the equation for part a

We are given the height - equation $s=-2.7t^{2}+40t + 6.5$ and we want to find $t$ when $s = 10$. So we set up the quadratic equation $-2.7t^{2}+40t+6.5 = 10$, which simplifies to $-2.7t^{2}+40t - 3.5=0$. The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=-2.7$, $b = 40$, and $c=-3.5$.

Step2: Calculate the discriminant

First, calculate the discriminant $\Delta=b^{2}-4ac=(40)^{2}-4\times(-2.7)\times(-3.5)=1600 - 37.8 = 1562.2$.

Step3: Find the values of $t$

Then, $t=\frac{-40\pm\sqrt{1562.2}}{2\times(-2.7)}=\frac{-40\pm39.525}{-5.4}$. We get two solutions: $t_1=\frac{-40 + 39.525}{-5.4}=\frac{-0.475}{-5.4}\approx0.09$ and $t_2=\frac{-40 - 39.525}{-5.4}=\frac{-79.525}{-5.4}\approx14.73$.

Step4: Set up the equation for part b

When the ball hits the moon's surface, $s = 0$. So we set up the quadratic equation $-2.7t^{2}+40t+6.5 = 0$. Using the quadratic formula with $a=-2.7$, $b = 40$, and $c = 6.5$.

Step5: Calculate the discriminant for part b

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

Step6: Find the value of $t$ for part b

$t=\frac{-40\pm\sqrt{1670.2}}{2\times(-2.7)}=\frac{-40\pm40.87}{-5.4}$. We take the positive root since time cannot be negative. $t=\frac{-40 + 40.87}{-5.4}$ is negative, and $t=\frac{-40-40.87}{-5.4}=\frac{-80.87}{-5.4}\approx14.98$.

Answer:

a. $0.09,14.73$ b. $14.98$