Question 825404
{{{ s(t) = -16t^2 + 80t + 44 }}}
(a)
{{{ s(t) = -16t^2 + 80t + 44 }}}
{{{ s(2) = -64 + 160 + 44
{{{ s(2) = 140 }}}
In 2 sec, the ball is 140 ft high
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(b)
The time , {{{t}}} when the 
height is maximum is
{{{ -b/(2a) }}} where
{{{ a = -165 }}}
{{{ b = 80 }}}
{{{ -b/(2a) = -80 / (2*(-16)) }}}
{{{ -b/(2a) = 80/32 }}}
{{{ -b/(2a) = 5/2 }}}
Plug this value back into equation to find {{{ s(2.5) }}}
{{{ s(2.5) = -16*(2.5)^2 + 80*2.5 + 44 }}} 
{{{ s(2.5) = -100 + 200 + 44 }}}
{{{ s(2.5) = 144 }}}
The maximum height is 144 ft
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(c)
The ball hits the ground when {{{ s(t) }}} is zero for
the 2nd time. Find the roots of the equation
{{{ s(t) = -16t^2 + 80t + 44 }}}
{{{ s(t) = 0 }}}
{{{ -16t^2 + 80t + 44 = 0 }}}
{{{ -4t^2 + 20t + 11 = 0 }}}
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = -4 }}}
{{{ b = 20 }}}
{{{ c = 11 }}}
{{{ t = ( -20 +- sqrt( 20^2 - 4*(-4)*11 )) / (2*(-4)) }}}
{{{ t = ( -20 +- sqrt( 400 + 176 )) / (-8) }}}
{{{ t = ( -20 +- sqrt( 576 )) / (-8) }}}
{{{ t = ( -20 - 24) / (-8) }}} ( I can't use the positive square root )
{{{ t = 11/2 }}}
The ball hits the ground in 5.5 sec
Here's the plot:
{{{ graph( 400, 400, -1, 6, -10, 160, -16x^2 + 80x + 44 ) }}}