Question 926546
{{{ y = -16x^2 + 44x + 30 }}}
If {{{ y }}} is feet and {{{ x }}} is seconds,
The object is fired up at {{{ x = 0 }}}, so
the initial height is:
{{{ y(0) = -16x^2 + 44*0 + 30 }}}
{{{ y(0) = 30 }}}
(a) 30 ft
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When the form of the equation is:
{{{ y = a*x^2 + b*x + c }}}, the time of the
maximum height is:
{{{ x[max] = -b/(2a) }}}
{{{ -b/(2a) = -44/( 2*(-16)) }}}
{{{ -b/(2a) = 11/8 }}}
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To find {{{ y[max] }}}:
{{{ y[max] = -16*(11/8)^2 + 44*(11/8) + 30 }}}
{{{ y[max] = -16*( 121/64 ) + 121/2 + 30 }}}
{{{ y[max] = -30.25 + 60.5 + 30 }}}
{{{ y[max] = 60.25 }}} 
(b) 60.25 ft
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The object is in the air until it hits the ground when
{{{ y = 0 }}}
{{{ 0 = -16x^2 + 44x + 30 }}}
{{{ -8x^2 + 22x + 15 = 0 }}}
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use quadratic formula:
{{{ x = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ x = (-22 +- sqrt( 22^2 - 4*(-8)*15 )) / (2*(-8)) }}}
{{{ x = (-22 +- sqrt( 484 + 480)) / (-16) }}}
{{{ x = ( -22 + sqrt( 964 ) ) / (-16) }}}
{{{ x = ( -22 - 31.048 ) / (-16) }}}
{{{ x = -53.048 / (-16) }}}
{{{ x = 3.3155 }}}
(c) 3.3155 sec
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check:
Here's the plot:
{{{ graph( 400, 400, -1, 5, -10, 70, -16x^2 + 44x + 30 ) }}}