Question 882146
The equation is a parabola
{{{ h(t) = -5t^2 + 22t + 7.8 }}}
The ball is released at {{{ t = 0 }}}
and {{{ h = 7.8 }}}
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You want to know what {{{ t }}} is
when {{{ h = 0 }}} for the 2nd time
( the ball comes back to earth )
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This is the same as finding both of
the roots of the equation
Set {{{ h(t) = 0 }}} and use the
quadratic formula to find roots
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{{{ -5t^2 + 22t + 7.8 = 0 }}}
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = -5 }}}
{{{ b = 22 }}}
{{{ c = 7.8 }}}
{{{ t = ( -22 +- sqrt( 22^2 - 4*(-5)*7.8 )) / (2*(-5)) }}}
{{{ t = ( -22 +- sqrt( 484 + 156 )) / (-10) }}}
{{{ t = ( -22 +- sqrt( 640 )) / (-10) }}}
{{{ t = ( -22 - 8*sqrt(10) ) / (-10) }}}
{{{ t = 2.2 + .8*sqrt(10) }}}
{{{ t = 2.2 + .8*3.1623 }}}
{{{ t = 2.2 + 2.53 }}}
{{{ t = 4.73 }}}
The ball lands in 4.73 sec
Note that I chose the largest positive root as the
correct one for when the ball will land
Here's the plot:
{{{ graph( 400, 400, -1, 6, -5, 35, -5x^2 + 22x + 7.8 ) }}}
{{{ t }}} at {{{ h = 0 }}} looks pretty close to {{{ 4.73 }}}