Question 1030625
Equation for vertical height, {{{ y }}}
{{{ y(t) = -16t^2 + 32t + 20 }}}
It looks like the diving board is {{{ 20 }}} ft
above the water, because when {{{ t=0 }}}
{{{ y(0) = -16*0^2 + 32*0 + 20 }}}
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You want to know what {{{ t }}} is when
{{{ y(t) = 0 }}} ( the water level )
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{{{ 0 = -16t^2 + 32t + 20 }}}
{{{ -16t^2 + 32t = -20 }}}
{{{ t^2 - 2t = 5/4 }}}
Complete the square
{{{ t^2 - 2t + (-2/2)^2 = 5/4 + (-2/2)^2 }}}
{{{ t^2 - 2t + 1 = 5/4 + 4/4 }}}
{{{ ( t - 1 )^2 = 9/4 }}}
{{{ ( t - 1 )^2  = (3/2)^2 }}}
{{{ t - 1 = 3/2 }}}
{{{ t = 5/2 }}}
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Equation for horizontal distance:
{{{ x(t) = 2t }}}
{{{ x( 5/2 ) = 2*( 5/2 ) }}}
{{{ x( 5/2 ) = 5 }}} ft
The diver travels 5 ft horizontally as he hits the water
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check:
{{{ y(t) = -16t^2 + 32t + 20 }}}
{{{ y(5/2) = -16*(5/2)^2 + 32*(5/2) + 20 }}}
{{{ y(5/2) = -16*( 25/4 ) + 160/2 + 20 }}}
{{{ y(5/2) = -100 + 80 + 20 }}}
{{{ y(5/2) = 0 }}} ( water level )
OK