An astronaut on the moon throws a ball upward. The astronaut releases the ball at a height of 6 feet (ho). with an initial velocity (vo) of 20 feet/second. The height (h) of the ball in feet is given by the equation:
h(t) = 2.7t^2 + vot + ho
Note: t represents the number of seconds after the ball was thrown.
a) What is the amount of time it takes the ball to achieve a height of 25 feet above the moon's surface?
b) At what height will the ball be after 5 seconds?
c) How long does it take for the ball to return to the surface of the moon?
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1 answer

don't forget your Algebra I now that you're taking calculus ...
(a) just solve for t in 2.7t^2 + 20t + 6 = 25
(b) just find h(5)
(c) solve for t in 2.7t^2 + 20t + 6 = 0 👍
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answered by oobleck
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