Question 925393: A rock is thrown upward with a velocity of 12 meters per second from the top of a 32 meter high cliff, and it misses the cliff on the way back down. When will the rock be 9 meters from the water, below?
I used this quadratic equation, −4.9t^2+2t+32=9 but wasn't sure if it was correct. I got the answer, 3.71. Is that the correct answer? Thanks for taking the time to help me!
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! use the following formula
s(t) = -4.9t^2 + v0t + s0 where vo is velocity and so is the height above ground when thrown, then we have
9 = -4.9t^2 + 12t +32
-4.9t^2 +12t + 23 = 0
use quadratic formula to solve for t
t = (-12 + square root(12^2 -4*(-4.9)*23) / (2*(-4.9)) = −1.264134887
t = (-12 - square root(12^2 -4*(-4.9)*23) / (2*(-4.9)) = 3.713114479
use t = 3.713114479 approx 3.71 meters
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