SOLUTION: I need help with this quadratic equation please: When a ball is thrown, its height in meters h after t seconds is given by the equation h= vt-5t^2 where v is the initial upwards v

Question 204879: I need help with this quadratic equation please:
When a ball is thrown, its height in meters h after t seconds is given by the equation h= vt-5t^2 where v is the initial upwards velocity in meters per second. If meters per second, find all values of t for which h=1 meter. Do not round any intermediate steps. Round the answer to 2 decimal places.
My try: Not even sure where to start..
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
When a ball is thrown, its height in meters h after t seconds is given by the equation h= vt-5t^2 where v is the initial upwards velocity in meters per second. If meters per second, find all values of t for which h=1 meter. Do not round any intermediate steps. Round the answer to 2 decimal places.
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h= vt-5t^2
Set h = 1
vt-5t^2 = 1
5t^2 - vt + 1 = 0
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The quadratic will have 2 real solutions, one for the ball going up, the other descending.
t = (v � sqrt(v^2 - 20))/10
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No value was given for v, so that's all that can be done.
If you plug in a value for v, you can solve for the 2 values of t.
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