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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-> 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
-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.
