# SOLUTION: A projectile is thrown upward so that its distance above the ground after t seconds is h=-15t(squared)+390t. After how many seconds does it reach its maximum height?

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A projectile is thrown upward so that its distance above the ground after t seconds is h=-15t(squared)+390t. After how many seconds does it reach its maximum height?      Log On

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 Question 430208: A projectile is thrown upward so that its distance above the ground after t seconds is h=-15t(squared)+390t. After how many seconds does it reach its maximum height?Found 3 solutions by Gogonati, poliphob3.14, ewatrrr:Answer by Gogonati(825)   (Show Source): You can put this solution on YOUR website!Solution:The projectile reach its maximum height when its velocity become zero. The velocity is:, for v=0 second. Answer: After 33 s the projectile reach maximum height. Done. Answer by poliphob3.14(115)   (Show Source): You can put this solution on YOUR website!Solution:The projectile reach its maximum height when its velocity become zero. The velocity is:, for v=0 second. Answer: After 33 s the projectile reach maximum height. Done. Answer by ewatrrr(11176)   (Show Source): You can put this solution on YOUR website! ``` Hi h = -15t^2+390t |parabola opening downward h = -15[(t-13)^2 -169] |completing square to find the Vertex h = -15(t-13)^2 +2535 In 13 sec it reaches it's maximum height ```