Question 801192
When you use feet and seconds, the 1st term
is {{{ -16t^2 }}} which is the effect of gravity
on the ball. The height above ground is a constant term
because if you set {{{ t = 0 }}} you get:
{{{ h(0) = -16*0^2 + 50*0 + 5 }}}
{{{ h(0) = 5 }}} ft, which it should be
-------------------------------
{{{ h(t) = -16t^2 + 50t + 5 }}}
You have to find the 2nd time that {{{ h(t) = 0 }}}
which is when the ball hits the ground.
{{{ h(t) = -16t^2 + 50t + 5 }}}
{{{ -16t^2 + 50t + 5 = 0 }}}
Use the quadratic formula
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = -16 }}}
{{{ b = 50 }}}
{{{ c = 5 }}}
{{{ t = ( -50 +- sqrt( 50^2 - 4*(-16)*5 )) / (2*(-16)) }}}
{{{ t = ( -50 +- sqrt( 2500 + 320 )) / ( -32 ) ) }}}
{{{ t = ( -50 +- sqrt( 2820 )) / ( -32 ) }}}
{{{ t = ( -50 - 53.104 ) / ( -32 ) }}}
{{{ t = 103.104 / 32 }}}
{{{ t = 3.222 }}}
The ball is in the air 3.222 sec
--------------------------
Here's a plot:
{{{ graph( 400, 400, -1, 4, -10, 50, -16x^2 + 50x + 5 ) }}}
--------------
The t-coordinate of maximum height is at {{{ -b/(2a)  = -50/(-32) }}}
{{{ t[max] = 1.563 }}} 
Plug this back into the equation to get {{{ h[max] }}}
--------------------
Another way is to find the {{{ t }}} value midway between
{{{ h(t) = 0 }}} and {{{ h( 3.222) = 0 , and then
find the {{{ h }}} value
------------------
Another way is just to make the plot