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  -> Quadratic Equation Customizable Word Problems -> 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


   

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) About Me  (Show Source):
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Solution:The projectile reach its maximum height when its velocity become zero.
The velocity is:+d%28h%29%2Fdt=v=-30t%2B390, for v=0
-30t%2B390=0
30t=390
t=33second.
Answer: After 33 s the projectile reach maximum height.
Done.

Answer by poliphob3.14(115) About Me  (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:+d%28h%29%2Fdt=v=-30t%2B390, for v=0
-30t%2B390=0
30t=390
t=33second.
Answer: After 33 s the projectile reach maximum height.
Done.

Answer by ewatrrr(24785) About Me  (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