SOLUTION: Problem: A rock is thrown from the top of a tall building. The distance, in feet, between the rock and the ground (T) seconds after it is thrown is given by d=-16Tsquared-2T+369. H

Question 308561: Problem: A rock is thrown from the top of a tall building. The distance, in feet, between the rock and the ground (T) seconds after it is thrown is given by d=-16Tsquared-2T+369. How long after the rock is thrown is it 330 feet from the ground?
I wrote the equation over again and replaced d as 330, making it 330=-16Tsquared-2T+369. I don't even know if this is what I'm supposed to be doing!
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Problem: A rock is thrown from the top of a tall building. The distance, in feet, between the rock and the ground (T) seconds after it is thrown is given by d=-16Tsquared-2T+369. How long after the rock is thrown is it 330 feet from the ground?
I wrote the equation over again and replaced d as 330, making it 330=-16Tsquared-2T+369. I don't even know if this is what I'm supposed to be doing!
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That's the right approach. Just solve for t
330 = -16t^2 - 2t + 369
-16t^2 - 2t + 39 = 0
(8t + 13)*(-2t + 3) = 0
t = -13/8 (Ignore it)
t = 3/2 seconds
It takes it 1.5 seconds to fall to 330 feet.

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