Question 1207285


{{{h = -16t^2 + 270t}}}

Now, we have to find out the time when {{{h = 390}}} feet.

{{{390 = -16t^2 + 270t}}}

{{{390 +16t^2 - 270t=0}}}

{{{16t^2 - 270t+390 =0}}}

{{{t=(-(-270)+-sqrt((-270)^2-4*16*390))/(2*16)}}}

{{{t=(270+-sqrt(47940))/32}}}

{{{t=(270+-218.95)/32}}}


{{{t=1.6}}} or {{{t=15.3}}}


answer is: 
 the rocket will be {{{390}}} feet above the ground

in {{{t=1.6}}} seconds on the way up, and

in {{{t=15.3}}} seconds on the way down