SOLUTION: An astronaut on the moon throws a baseball upward. The height h of the ball, in feet, x seconds after the ball is thrown is given by the equation.
h+-2.7x^2+30x+6.5
After ho
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h+-2.7x^2+30x+6.5
After ho
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Question 130147: An astronaut on the moon throws a baseball upward. The height h of the ball, in feet, x seconds after the ball is thrown is given by the equation.
h+-2.7x^2+30x+6.5
After how many seconds is the ball 12 feet above the moon's surface? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The height h of the ball, in feet, x seconds after the ball is thrown is given by the equation.
h(t)= -2.7x^2+30x+6.5
After how many seconds is the ball 12 feet above the moon's surface?
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Let the height be 12 and solve for "x".
-2.7x^2 + 30x + 6.5 = 12
-2.7x^2 + 30x - 4.5 = 0
x = [-30 +- sqrt(900 -4*-2.7*-4.5)]/-5.4
x = [-30 +- sqrt(851.4)]/-5.4
x = [-30 +- 29.17876]/-5.4
x = 10.96 seconds and x = 0.1521 seconds
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Cheers,
Stan H.
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