SOLUTION: AN Astronaut standing on the surface of the moon throws a rock into the air with and initial velocity of 27 feet per second. the astronaut hand is 6 feet above the surface of the m

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Question 860213: AN Astronaut standing on the surface of the moon throws a rock into the air with and initial velocity of 27 feet per second. the astronaut hand is 6 feet above the surface of the moo. The height of the rock is given by h= -2.7f2+27t+6.
How many seconds was the rock in the air?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
AN Astronaut standing on the surface of the moon throws a rock into the air with and initial velocity of 27 feet per second.
the astronaut hand is 6 feet above the surface of the moon.
The height of the rock is given by h= -2.7f2+27t+6.
How many seconds was the rock in the air?
:
When the rock strikes the ground, h=0
-2.7t^2 + 27t + 6 = 0
Use the quadratic formula to find t,
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
where a=-2.7, b=27, c=6
t = t+=+%28-27+%2B-+sqrt%2827%5E2-4%2A-2.7%2A6+%29%29%2F%282%2A-2.7%29+
do the math, I got a positive solution of t=10.275 sec, see what you get.