Question

a projectile is thrown upward so that its distance above the ground after t seconds is h(t)=-15t^2+390t. after how many seconds does it reach its maximum height?

Answer 1

A lot

Answer 2

get the first derivative and equal it to zero
-30t+390=0
t= 13 sec

Answer 3

@Kellbell456 did you get it

Answer 4

good luck with that program...

Answer 5

@satellite73 what do you mean by program??

Answer 6

First you need to determine what the maximum height is.

Answer 7

i bet $7 this is not a calculus class
first coordinate of the vertex is \(-\frac{b}{2a}\) is what is asked for here

Answer 8

\[h(t)=-15t^2+390t\]
\[-\frac{b}{2a}=-\frac{390}{2\times (-15)}\] etc

Answer 9

equation is wrong in any case but no matter

Answer 10

13 looks good to me

Answer 11

me too

Answer 12

Thank you satellite73. :)