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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

  1. 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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