Question 1022423
{{{ h(t) = -16t^2 + 80t + 4 }}}
{{{ h(t) }}} is the height at whatever time {{{ t }}} is
{{{ -16t^2 }}} is due to downward pull of gravity
{{{ 80t }}} is upward velocity
{{{ 4 }}} is initial height
------------------------
You want to find when {{{ h(t) = 0 }}}, because that
is the ground level
{{{ 0 = -16t^2 + 80t + 4 }}}
{{{ 0 = -4t^2 + 20t + 1 }}}
Use quadratic formula
{{{ t = ( -b +- sqrt( b^2-4*a*c )) / (2*a) }}}
{{{ a = -4 }}}
{{{ b = 20 }}}
{{{ c = 1 }}}
{{{ t = ( -20 +- sqrt( 20^2 - 4*(-4)*1 )) / (2*(-4)) }}}
{{{ t = ( -20 +- sqrt( 400 + 16 )) / (-8) }}}
{{{ t = ( -20 +- sqrt( 416 )) / (-8) }}}
{{{ t = ( -20 - 20.396 ) / (-8) }}}
( note that I used the negative square root so
that I would end up with positive time )
{{{ t = ( -40.396 ) / (-8) }}}
{{{ t = 5.05}}} sec
The ball hits ground in 5.05 sec
Here's the plot:
{{{ graph( 400, 400, -1, 6, -10, 120, -16x^2 + 80x + 4 ) }}}
Hope I got it!