Question 660330
In words:
You need an equation that has 3 parts
to describe the height above ground:
h = ( height change due to throwing ) - ( height change due to gravity )
+ ( the initial height above ground )
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Actual formula looks like:
{{{ h = v[0]*t - 16t^2 + h[0] }}}
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(a)
The first object is launched downward at 48ft/sec. 
That means the {{{ v[0]*t }}} term is minus, the
same as gravity term
{{{ h = -v[0]*t - 16t^2 + h[0] }}}
{{{ h[0] = 160 }}} ft
{{{ v[0] = 48 }}} ft/sec downward
{{{ h = -48t - 16t^2 + 160 }}}
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(b)
The second object is dropped.
That means the {{{ v[0]*t }}} term is zero ( no throwing )
{{{ h = -16t^2 + 160 }}}
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(c)
The third object is launched upward at 96ft/sec.
The {{{ v[0]*t }}} term is positive
{{{ h = 96t - 16t^2 + 160 }}}
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determine the length of time the third object was in the air. 
That means the object is trhown up, reaches a peak, and
comes back to the ground
Note that when {{{ t = 0 }}}, the equation becomes
{{{ h = 160 }}} which means object hasn't been thrown yet
You want to find {{{ t }}} when {{{ h = 0 }}} ( back to ground )
{{{ 0 = -16t^2 + 96t + 160 }}}
Divide both sides by {{{ 16 }}}
{{{ 0 = -t^2 + 6t + 10 }}}
Use quadratic formula
{{{ t = ( -b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = -1 }}}
{{{ b = 6 }}}
{{{ c = 10 }}}
{{{ t = ( -6 +- sqrt( 36 - 4*(-1)*10 ))/(2*(-1)) }}} 
{{{ t = ( -6 +- sqrt( 36 + 40 ))/(-2) }}} 
{{{ t = ( -6 +- sqrt( 76 ))/(-2) }}} 
{{{ t = ( -6 - sqrt( 76 )) / (-2) }}}
This gives the time in the air
Here's the plot of the equation {{{ h = -16t^2 + 96t + 160 }}}
You can see the height of {{{ 160 }}} ft where object is thrown
up and the time of 7+ sec when it hits ground
{{{ graph( 400, 400, -2, 10, -20, 320, -16x^2 + 96x + 160 ) }}}