Question 946580
{{{ g(x) = -.04x^2 + 2.1x + 6.1 }}}
This is in the form:
{{{ g(x) = a*x^2 + b*x + c }}} where
{{{ a = -.04 }}}
{{{ b = 2.1 }}}
{{{ c = 6.1 }}}
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(a)
The {{{ x }}} component of the maximum
height is :
{{{ x[max] = -b/(2a) }}}
{{{ x[max] = -(2.1)/(2*(-.04)) }}}
{{{ x[max] = 2.1 / .08 }}}
{{{ x[max] = 26.25 }}}
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{{{ g[max] = -.04*( 26.25 )^2 + 2.1*26.25 + 6.1 }}}
{{{ g[max] = -27.5625 + 55.125 + 6.1 }}}
{{{ g[max] = 33.6625 }}}
To the nearest 10th, the maximum height
is 33.7 ft
The maximum height is reached 26.25 ft
from the point of release
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(b)
{{{ g(x) = -.04x^2 + 2.1x + 6.1 }}}
{{{ -.04x^2 + 2.1x + 6.1 = 0 }}}
Use the quadratic formula
{{{ x = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = -.04 }}}
{{{ b = 2.1 }}}
{{{ c = 6.1 }}}
{{{ x = ( -2.1 +- sqrt( 2.1^2 - 4*(-.04)*6.1 )) / (2*(-.04)) }}}
{{{ x = ( -2.1 +- sqrt( 4.41 + .976)) / (-.08) }}}
{{{ x = ( -2.1 - sqrt( 5.386 )) / ( -.08 ) }}}
{{{ x = ( -2.1 - 2.32078 ) / ( -.08 ) }}}
{{{ x = ( -4.42078 ) / ( -.08 ) }}}
{{{ x = 55.2596 }}}
The maximum horizontal distance is:
55.3 ft to the nearest tenth
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(c)
To find height from which shot is released, se the
horizontal distance, {{{ x }}} to {{{ 0 }}}
{{{ g(0) = -.04*0^2 + 2.1*0 + 6.1 }}}
{{{ g(0) = 6.1 }}}
The shot was released from a height of 6.1 ft
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Here's a plot:
{{{ g(x) = -.04x^2 + 2.1x + 6.1 }}}
{{{ graph( 400, 400, -6, 60, -4, 40, -.04x^2 + 2.1x + 6.1  ) }}}