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 30 ft per sec. the height s of the ball in feet is given by the equation $s = -2.7t^2 + 30t + 6.5$, where t is the number of seconds after the ball was thrown. complete parts a and b.\
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a. after how many seconds is the ball 12 ft above the moons surface?\
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after \\(\square\\) 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.)

Explanation:

Step1: Set up the equation

We know the height equation is \( s = -2.7t^{2}+30t + 6.5 \), and we want to find \( t \) when \( s = 12 \). So we set up the equation:
\( 12=-2.7t^{2}+30t + 6.5 \)

Step2: Rearrange to standard quadratic form

Subtract 12 from both sides to get:
\( 0=-2.7t^{2}+30t + 6.5 - 12 \)
Simplify the right - hand side:
\( 2.7t^{2}-30t + 5.5 = 0 \)
Here, \( a = 2.7 \), \( b=-30 \), \( c = 5.5 \)

Step3: Use the quadratic formula

The quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)
First, calculate the discriminant \( D=b^{2}-4ac \)
\( D=(-30)^{2}-4\times2.7\times5.5 \)
\( D = 900-59.4=840.6 \)
Then, \( t=\frac{30\pm\sqrt{840.6}}{2\times2.7}=\frac{30\pm28.99}{5.4} \)

Step4: Calculate the two solutions

For the plus sign:
\( t_1=\frac{30 + 28.99}{5.4}=\frac{58.99}{5.4}\approx10.92 \)
For the minus sign:
\( t_2=\frac{30 - 28.99}{5.4}=\frac{1.01}{5.4}\approx0.19 \)

Answer:

0.19, 10.92