An astronaut on the moon throws a baseball upward. The astronaut is 6.5 feet tall, the velocity is 40 feet per second. How many seconds is the ball 18 feet above the moon? And how many seconds will it take for the ball to hit the moons surface?

Step-by-step explanation:

Gravity on Earth is -32 ft/s².  Gravity on the moon is 1/6 of that (-5.33 ft/s²).

The initial height is 6.5 ft, and the initial velocity is 40 ft/s.

So the height of the ball after t seconds is:

y = ½ a t² + v t + h

y = ½ (-5.33) t² + (40) t + (6.5)

y = -2.67t² + 40t + 6.5

When y = 18:

18 = -2.67t² + 40t + 6.5

2.67t² − 40t + 11.5 = 0

t = [ -(-40) ± √((-40)² − 4(2.67)(11.5)) ] / 2(2.67)

t = 0.293 or 14.707

The ball is 18 feet above the ground after 0.293 seconds and 14.707 seconds.

When y = 0:

0 = -2.67t² + 40t + 6.5

2.67t² − 40t − 6.5 = 0

t = [ -(-40) ± √((-40)² − 4(2.67)(-6.5)) ] / 2(2.67)

t = -0.161 or 15.161

t can't be negative, so the ball lands after 15.161 seconds.

answer: c

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