Question 1180599
6 feet, the constant in the quadratic.
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The maximum height occurs at d=-b/2a or -3/-0.4 or 7.5 
h(7.5)=17.25 feet.
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set h(d)=10
so -0.2d^2+3d+6=10
and 0.2d^2-3d+4=0
d^2-15d+20=0, multiplying by 5
d=(1/2)(15+/- sqrt(225-80)); sqrt 145=12.04
the distance will be 1.48 feet and 13.52 feet
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There are several possible solutions since the ball doesn't have to be dead center to go in, but since there isn't information about that, we can still figure the optimum distance. That would be when the quadratic equals 10 above, so the distance would be 13.52 feet. The other solution would require the ball to be traveling upward through the net and that would not be allowable.  
Remember that the 17 foot distance where the ball lands on the floor would only be if it didn't go through the net.

{{{graph(300,300,-10,20,-10,20,10,-0.2x^2+3x+6)}}}