Question 285567
When {{{t = 0}}}, the rocket is still on the ground, and
{{{h = 50}}} (height of rocket?)
The rocket is also on the ground when it comes 
back down at {{{h = 0}}}, so
{{{-16t^2 + 280t + 50 = 0}}}
Using quadratic equation:
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = -16}}}
{{{b = 280}}}
{{{c = 50}}}
{{{t = (-280 +- sqrt( 280^2-4*(-16)*50 ))/(2*(-16)) }}}
{{{t = (-280 +- sqrt( 78400 + 64*50 ))/(-32)) }}}
{{{t = (-280 +- sqrt( 78400 + 3200 ))/(-32)) }}}
{{{t = (-280 +- sqrt( 81600 ))/(-32)) }}}
{{{t = (-280 +- 285.657)/(-32)) }}}
The (+) square root gives me negative {{{t}}},
 so I can't use it
{{{t = (-565.657)/(-32)) }}}
{{{t = 17.677}}} sec
The rocket comes back to the ground in 17.677sec
Here's a plot of {{{h}}} vs {{{t}}} also:
{{{ graph( 500, 500, -5, 22, -50, 1300, -16x^2 + 280x + 50) }}}