The height of a ball thrown in the air is given by $h(x) = \frac {–1}{12} x^2 + 6x+ 3$, where x is the horizontal distance in feet from the point at which the ball is thrown.
c. How far from the thrower does the ball strike the ground?
The height of a ball thrown in the air is given by $h(x) = \frac {–1}{12} x^2 + 6x+ 3$, where x is the horizontal distance in feet from the point at which the ball is thrown.
c. How far from the thrower does the ball strike the ground?
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The striker is at the point $(0,0)$ and the ball comes down at another point $(x,0)$. We need to calculate the $x$ (the horizontal distance).
This is done by setting $h(x)=0 \Leftrightarrow -\frac{1}{12}x^2+6x+3=0$.
$-\frac{1}{12}x^2+6x+3=0$
$x^2-72x-36=0$
Quadratic formula: $x_{1,2}=\frac{72\pm\sqrt{(-72)^2-4\cdot1\cdot-36}}{2\cdot1}=36\pm6\sqrt{37}$.
$x=36+6\sqrt{37}\approx72.5$ is what we're looking for, so the horizontal distance to the point where the ball comes down is $72.5$ feet.