Question 763698
{{{ h(t) = -16t^2+vt+s }}}
notice I replaced {{{ h }}} with {{{ h(t) }}}
It's just a matter of preference.
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Think of the 1st term as how the force
of gravity is affecting the height.
The other terms have nothing to do with gravity.
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The 2nd term has to do with how fast you
release the ball, and whether you throw it
up or down (+) is for up, (-) is for down.
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The 3rd term is only the height above ground
from which the ball is released.
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It's important to see that at {{{ t = 0 }}}, just
as the ball is released, 
{{{ h(0) = -16*0^2+v*0+s }}} 
{{{ h(0) = s }}}, the height from which ball is released,
and you are given that {{{ s = 100 }}}
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You are asked to find {{{ t }}} when {{{ h(t) = 200 }}} ft
Note that the positive {{{ 96 }}} ft/sec has to fight 
against the negative {{{ 16 }}} ft/sec2 which is acceleration
due to gravity.
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{{{ 200 = -16t^2 + 96t + 100 }}}
{{{ -16t^2 + 96t - 100 = 0 }}}
{{{ -4t^2 + 24t - 25 = 0 }}}
use quadratic formula
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = -4 }}}
{{{ b = 24 }}}
{{{ c = -25 }}}
{{{ t = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ t = (-24 +- sqrt( 24^2 - 4*(-4)*(-25) )) / (2*(-4)) }}} 
{{{ t = (-24 +- sqrt( 576 - 400 )) / (-8) ) }}} 
{{{ t = (-24 +- sqrt( 176 )) / (-8) ) }}} 
{{{ t = (-24 +- 13.2665 ) / (-8) ) }}}
There are 2 valid solutions. There is a height when 
the ball is on the way up to it's peak, and also a
height when the ball is on the way down from it's peak.

{{{ t = ( -24 + 13.2665 ) / (-8) }}} 
{{{ t = 1.3417 }}} sec
and, also
{{{ t = ( -24 - 13.2665 ) / (-8) }}}
{{{ t = -37.2665 / (-8) }}}
{{{ t = 4.6583 }}} sec
( note that you should never end up with a negative time )
Here's a plot:
{{{ graph( 400, 400, -2, 20, -20, 250, -16x^2 + 96x + 100 ) }}}
At t = 1.34 sec the height is 200 ft and
At t = 4.66 sec the height is 200 ft
You can see this on plot
Hope this helps