A golfer hits a ball with an initial speed of v ft/sv\text{ ft/s} at a 35°35\degree angle. The ball is 11 ft\text{ft} above the ground when hit and aims for a hole that is 450450 ft\text{ft} away. The height hh (in feet) of the ball after it has traveled a horizontal distance of xx ft\text{ft} is given by the function h(x)=25x2v2+x+1h(x)=\frac{$$25x^2$$}{$$v^2$$}^{}+x+1. If the diameter of the hole is 4.254.25 inches, what is the distance ss from the center of the golf ball to the edge of the hole when the ball has traveled xx feet? Neglect the size of the ball.

sedge=(450−x)2+(25x2v2+x+1)2−0.1771s_{\text{edge}}=\sqrt{$$(450-x)^2$$+\left(\frac{$$25x^2$$}{$$v^2$$}^{}+x+1\$$right)^2$$}-0.1771

(450+x)2+(25x2v2+x+1)2+0.1771\sqrt{$$(450+x)^2$$+\left(\frac{$$25x^2$$}{$$v^2$$}^{}+x+1\$$right)^2$$}+0.1771

sedge=(450−x)2+(25x2v2+x+1)2+0.1771s_{\text{edge}}=\sqrt{$$(450-x)^2$$+\left(\frac{$$25x^2$$}{$$v^2$$}^{}+x+1\$$right)^2$$}+0.1771

sedge=(450+x)2+(25x2v2+x+1)2−0.1771s_{\text{edge}}=\sqrt{$$(450+x)^2$$+\left(\frac{$$25x^2$$}{$$v^2$$}^{}+x+1\$$right)^2$$}-0.1771

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Problem 37Multiple Choice

A golfer hits a ball with an initial speed of $v ft/s$ at a $35 degrees$ angle. The ball is $1$ $ft$ above the ground when hit and aims for a hole that is $450$ $ft$ away. The height $h$ (in feet) of the ball after it has traveled a horizontal distance of $x$ $ft$ is given by the function $h (x) = {\frac{25 x^{2}}{v^{2}}}^{} + x + 1$ . If the diameter of the hole is $4.25$ inches, what is the distance $s$ from the center of the golf ball to the edge of the hole when the ball has traveled $x$ feet? Neglect the size of the ball.

$\sqrt{(450 + x)^{2} + {({\frac{25 x^{2}}{v^{2}}}^{} + x + 1)}^{2}} + 0.1771$

$s_{edge} = \sqrt{(450 - x)^{2} + {({\frac{25 x^{2}}{v^{2}}}^{} + x + 1)}^{2}} + 0.1771$

$s_{edge} = \sqrt{(450 + x)^{2} + {({\frac{25 x^{2}}{v^{2}}}^{} + x + 1)}^{2}} - 0.1771$

$s_{edge} = \sqrt{(450 - x)^{2} + {({\frac{25 x^{2}}{v^{2}}}^{} + x + 1)}^{2}} - 0.1771$

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