SOLUTION: An arrow is shot into the air is 144t - 4.9t2 meters above the ground t seconds after it is released. During what period(s) of time is the arrow above 68.6 meters? Round your answe

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Question 876037: An arrow is shot into the air is 144t - 4.9t2 meters above the ground t seconds after it is released. During what period(s) of time is the arrow above 68.6 meters? Round your answer to the nearest .01 second.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
t = .48sec and
68.6 = -4.9t^2 + 144t
-4.9t^2 + 144t - 68.6 = 0
t = .48 and t = 28.90
from t = .48sec to 28.90sec
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -4.9x%5E2%2B144x%2B-68.6+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28144%29%5E2-4%2A-4.9%2A-68.6=19391.44.

Discriminant d=19391.44 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-144%2B-sqrt%28+19391.44+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28144%29%2Bsqrt%28+19391.44+%29%29%2F2%5C-4.9+=+0.484372370178204
x%5B2%5D+=+%28-%28144%29-sqrt%28+19391.44+%29%29%2F2%5C-4.9+=+28.9033827318626

Quadratic expression -4.9x%5E2%2B144x%2B-68.6 can be factored:
-4.9x%5E2%2B144x%2B-68.6+=+-4.9%28x-0.484372370178204%29%2A%28x-28.9033827318626%29
Again, the answer is: 0.484372370178204, 28.9033827318626. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-4.9%2Ax%5E2%2B144%2Ax%2B-68.6+%29