Question 683994
How high was the ball when the shooter shot it?


{{{h= -0.2d^2 + 3d + 6}}}


{{{h= -0.2(0)^2 + 3(0) + 6}}} ... Plug in d = 0 (since the ball is zero ft away from the player when it is shot)


{{{h= -0.2(0) + 3(0) + 6 }}}


{{{h= 0 + 0 + 6 }}}


{{{h = 6}}}


So the ball is 6 ft off the ground when the shooter shot it.


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What was the maximum height obtained by the ball?


To answer this question, we need to find the vertex


First replace all 'h' terms with y. Then replace all 'd' terms with 'x'


Now find the x coordinate of the vertex


{{{x = -b/(2a)}}}


{{{x = -3/(2(-0.2))}}}


{{{x = 7.5}}}


Then plug this in to find the y coordinate of the vertex


{{{y = -0.2d^2 + 3d + 6}}}


{{{y = -0.2(7.5)^2 + 3(7.5) + 6}}}


{{{y = -0.2(56.25) + 3(7.5) + 6}}}


{{{y = -11.25 + 22.5 + 6}}}


{{{y = 17.25}}}


The y coordinate of the vertex represents the highest point.


So the max height is 17.25 ft.


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A rim of a basketball net is 10 feet high. For what horizontal distance was the ball above the rim of the basketball net? 



Plug in h = 10 and solve for d


{{{h = -0.2d^2 + 3d + 6}}}


{{{10 = -0.2d^2 + 3d + 6}}}


{{{0 = -0.2d^2 + 3d + 6 - 10}}}


{{{0 = -0.2d^2 + 3d - 4}}}

 
{{{-0.2d^2 + 3d - 4 = 0}}}


{{{10(-0.2d^2 + 3d - 4) = 10*0}}}


{{{-2d^2 + 30d - 40 = 0}}}


{{{-1(-2d^2 + 30d - 40) = -1*0}}}


{{{2d^2 - 30d + 40 = 0}}}


{{{2(d^2 - 15d + 20) = 0}}}


{{{d^2 - 15d + 20 = 0/2}}}


{{{d^2 - 15d + 20 = 0}}}


Now use the quadratic formula to solve for d


{{{d = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{d = (-(-15)+-sqrt((-15)^2-4(1)(20)))/(2(1))}}} Plug in {{{a = 1}}}, {{{b = -15}}}, {{{c = 20}}}


{{{d = (15+-sqrt(225-(80)))/(2)}}}


{{{d = (15+-sqrt(145))/2}}}


{{{d = (15+sqrt(145))/2}}} or {{{d = (15-sqrt(145))/2}}}


{{{d = 13.520797}}} or {{{d = 1.479203}}}


So the distance is either {{{d = 13.520797}}} or {{{d = 1.479203}}}, which means that the distance it travels about 13.520797 - 1.479203 = 12.041594 feet horizontally when the ball is above the 10 ft rim.