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?
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(855) (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(24785) (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