SOLUTION: A ball thrown from a height of 1 meter with an initial upward velocity of 25 m/s. The balls height h (in meters) after t seconds is given by the following h=1+25t-5t^2 . Find all v

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: A ball thrown from a height of 1 meter with an initial upward velocity of 25 m/s. The balls height h (in meters) after t seconds is given by the following h=1+25t-5t^2 . Find all v      Log On


   

Question 1144482: A ball thrown from a height of 1 meter with an initial upward velocity of 25 m/s. The balls height h (in meters) after t seconds is given by the following h=1+25t-5t^2 . Find all values of t for which the balls height is 11 meters. Round to the nearest hundredth
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
The ball is at the height 11 meters, when


    1 + 25t - 5t^2 = 11   meters.


Solve this quadratic equation to find "t".


    5t^2 - 25t + 10 = 0

      t^2 - 5t + 2 = 0

      t%5B1%2C2%5D = %285+%2B-+sqrt+%285%5E2+-+4%2A2%29%29%2F2 = %285+%2B-+sqrt%2817%29%29%2F2


      t%5B1%5D = %285+-+sqrt%2817%29%29%2F2 = 0.44 seconds (approximately),   and

      t%5B2%5D = %285+%2B+sqrt%2817%29%29%2F2 = 4.56 seconds (approximately).



   


    Plot y = 1+%2B+25x+-+5x%5E2 (red) and y = 11 (green)



Time moment t%5B1%5D  corresponds to the ball moving up.

Time moment t%5B2%5D  corresponds to the ball falling down.