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

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

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

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

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

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Problem 33Multiple 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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