Question 1028630
You are given that {{{ h(t) = 13 }}} where
{{{ h(t) = -5t^2 + 16t + 1 }}}
{{{ -5t^2 + 16t + 1 = 13 }}} 
{{{ -5t^2 + 16t - 12 = 0 }}}
Use quadratic formula
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = -5 }}}
{{{ b = 16 }}}
{{{ c = -12 }}}
{{{ t = ( -16 +- sqrt( 16^2 - 4*(-5)*(-12) )) / (2*(-5) ) }}}
{{{ t = ( -16 +- sqrt( 256 - 240 )) / (-10 ) }}}
{{{ t = ( -16 +- sqrt( 16 )) / (-10 ) }}}
{{{ t = ( -16 + 4 ) / (-10) }}}
{{{ t = (-12)/(-10) }}}
{{{ t = 1.2 }}}
and also
{{{ t = ( -16 - 4 ) / (-10) }}}
{{{ t = (-20)/(-10) }}}
{{{ t = 2 }}}
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The first time the height of the ball is at 13 m
is 1.2 sec
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check:
Here's the plot:
{{{ graph( 600, 600, -2, 4, -2, 15, -5x^2 + 16x + 1 ) }}}