SOLUTION: A cannon ball is thrown into the air from a height ,(h),,of 12 m. The height of the ball above ground t seconds after being shot is approximated by h(t) = -5t squared +15 t +12. De

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Question 1117051: A cannon ball is thrown into the air from a height ,(h),,of 12 m. The height of the ball above ground t seconds after being shot is approximated by h(t) = -5t squared +15 t +12. Determine when the ball is 22 m above ground. Thanks
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A cannon ball is thrown into the air from a height ,(h),,of 12 m. The height of the ball above ground t seconds after being shot is approximated by h(t) = -5t squared +15 t +12. Determine when the ball is 22 m above ground
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h(t) = -5t^2 + 15t + 12 = 22
It's a quadratic.
Solve for t
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
-5t^2 + 15t + 12 = 22


-5t^2 + 15t + (12-22) = 0


-5t^2 + 15t - 10 = 0


Discriminant of the equation  d = 15^2 - 4*(-5)*(-10) = 225 - 200 = 25


 =  = .


The two roots are  


 =  =  = 1   and


 =  =  = 2.


Answer.   = 1 second on the way up,   and   = 2 second on the way down.





Plot y =  (red) and y = 22 (green)


You consider the plot at t >= 0.


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