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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 isgiven 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.a. After how many seconds is the ball 18 ft above the moon's surface?After ____ seconds the ball will be 18 ft above the moon's surface.(Round to the nearest hundredth as needed. Use a comma to separate answers as needed.)

Sagot :

[tex]s=-2.7t^2+30t+6.5[/tex]

In order to find when the ball will be 18 ft above the moon's surface, we need to equal the expression to 18

[tex]18=-2.7t^2+30t+6.5[/tex]

then, solve the associated quadratic expression

[tex]\begin{gathered} 0=-2.7t^2+30t+6.5-18 \\ 0=-2.7t^2+30-11.5 \\ using\text{ }the\text{ }quadratic\text{ }formula \\ x=\frac{-30\pm\sqrt{(30)^2-4\ast(-2.7)\ast(-11.5)}}{2\ast(-2.7)} \\ x_1\cong0.40 \\ x_2\cong10.72 \end{gathered}[/tex]

answer:

after 0.40 seconds the ball will be 18 ft above the surface

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