Answer to Question #42997 in Calculus for shasha

Question

A projectile is thrown upward so that its distance above the ground after t seconds is h = - 16t2 + 440t.

After how many seconds does it reach its maximum height?


help me please

Answer 1

The trajectory of its motion is a parabola with downward branches (the coefficient with t^2 less then 0)
The maximum height is in the vertex of this parabola. Using the formula t₀ = -b/2a we get
t_max = -440/(-2*16) = 13.75 seconds
Answer: 13.75 seconds.