SOLUTION: An object launched upward with initial velocity of 4.9 m/s from a height of 9.8 m will have a height of, S(t)=-4.9t^2+4.9t+9.8, where S is in meters and t is in seconds. a). How l

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Question 457645: An object launched upward with initial velocity of 4.9 m/s from a height of 9.8 m will have a height of, S(t)=-4.9t^2+4.9t+9.8, where S is in meters and t is in seconds.
a). How long will it take the object to hit the ground?
b). Find the interval on which the height of the object is greater than 2.8 m.
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
s=-4.9t2+4.9t+9.8
4.9t2-4.9t-9.8=0
t2-t-2=0
(t+1)(t-2)=0
t=2 or -1
Throwing out the negative answer, the object will hit the ground in 2 seconds.
2.8<-4.9t2+4.9t+9.8
0<-4.9t2+4.9t+7
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -4.9x%5E2%2B4.9x%2B7+=+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=%284.9%29%5E2-4%2A-4.9%2A7=161.21.

Discriminant d=161.21 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4.9%2B-sqrt%28+161.21+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284.9%29%2Bsqrt%28+161.21+%29%29%2F2%5C-4.9+=+-0.795596939086932
x%5B2%5D+=+%28-%284.9%29-sqrt%28+161.21+%29%29%2F2%5C-4.9+=+1.79559693908693

Quadratic expression -4.9x%5E2%2B4.9x%2B7 can be factored:
-4.9x%5E2%2B4.9x%2B7+=+-4.9%28x--0.795596939086932%29%2A%28x-1.79559693908693%29
Again, the answer is: -0.795596939086932, 1.79559693908693. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-4.9%2Ax%5E2%2B4.9%2Ax%2B7+%29

The area in the upper right quadrant in your graph represents the time the object is above 2.8 meters..