Question 426739
{{{ h(t) = -16t^2 + 40t }}}
The maximum height occurs at 
{{{ t[max] = -b/(2a) }}} where 
{{{a = -16}}}
{{{b = 40}}}
{{{ -b/(2a) = -40/(2*-16) }}}
{{{ t[max] = 5/4 }}}
To find the max height,
{{{ h(5/4) = -16*(5/4)^2 + 40*(5/4) }}}
{{{ h(5/4) = - 25 + 50 }}}
The max height is 25 ft
It will land when {{{h = 0}}}
{{{ 0 = -16t^2 + 40t }}}
{{{ 0 = 8t*(-2t + 5) }}}
There are 2 ways for h(t) to be zero, 
{{{ 8t = 0 }}}
{{{ t = 0 }}} ( at takeoff), and
{{{ -2t + 5 = 0}}}
{{{2t = 5}}}
{{{ t = 5/2 }}}
The rocket will land in 2.5 sec
check answer:
{{{ h(t) = -16t^2 + 40t }}}
{{{ 0 = -16*2.5^2 + 40*2.5 }}}
{{{ 0 = -16*6.25 + 100 }}}
{{{ 0 = -100 + 100 }}}
{{{ 0= 0 }}}
OK
Here's a plot:
{{{ graph( 400, 400, -1, 3, -1, 30, -16x^2 + 40x) }}}