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# phyhw5 - valdez(vv689 Homework 05 orin(58140 This print-out...

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valdez (vv689) – Homework 05 – florin – (58140) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The speed of an arrow fired from a compound bow is about 63 m / s. An archer sits astride his horse and launches an arrow into the air, elevating the bow at an angle of 46 above the horizontal and 2 . 5 m above the ground. The acceleration of gravity is 9 . 81 m / s 2 . 63 m / s 46 2 . 5 m range What is the arrow’s range? Assume: The ground is level. Ignore air resistance. Correct answer: 406 . 741 m. Explanation: Let : v o = 63 m / s , θ = 46 , h = 2 . 5 m , and g = 9 . 81 m / s 2 . v o θ h range In the projectile motion, we have x = ( v o cos θ ) t t = x v o cos θ horizontally, and y = h + ( v o sin θ ) t - 1 2 g t 2 vertically. Thus y = h + x v o sin θ v o cos θ - 1 2 g x 2 v 2 o cos 2 θ = 0 when the arrow lands. Thus 2 v 2 o h cos 2 θ + 2 x v 2 o sin θ cos θ - g x 2 = 0 - 2 v 2 o h cos 2 θ - x v 2 o sin 2 θ + g x 2 = 0 x = v 2 o sin 2 θ ± radicalBig v 4 o sin 2 2 θ + 8 g v 2 o h cos 2 θ 2 g . Since v 4 o sin 2 2 θ + 8 g v 2 o h cos 2 θ = (63 m / s) 4 (sin 2 92 ) + (8) (9 . 81 m / s 2 ) × (63 m / s) 2 (2 . 5 m) (cos 2 46 ) = 1 . 61095 × 10 7 m 2 / s 2 , then x = (63 m / s) 2 (sin 92 ) 2 (9 . 81 m / s 2 ) + radicalbig 1 . 61095 × 10 7 m 2 / s 2 2 (9 . 81 m / s 2 ) = 406 . 741 m . 002 (part 2 of 2) 10.0 points Now assume his horse is at full gallop, moving in the same direction as he will fire the arrow, and that he elevates the bow the same way as before. What is the range of the arrow at this time if the horse’s speed is 40 m / s? Correct answer: 778 . 503 m. Explanation: The arrow’s speed relative to the ground and the angle of elevation relative to the ground are changed v x = v arrow + v archer = (63 m / s) cos 46 + 40 m / s = 83 . 7635 m / s .

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valdez (vv689) – Homework 05 – florin – (58140) 2 v y = (63 m / s) sin 46 = 45 . 3184 m / s The arrow’s speed relative to the ground is then given by v o = radicalBig v 2 x + v 2 y = radicalBig (83 . 7635 m / s) 2 + (45 . 3184 m / s) 2 = 95 . 237 m / s , and the angle of elevation relative to the ground is θ = tan 1 v y v x = tan 1 parenleftbigg 45 . 3184 m / s 83 . 7635 m / s parenrightbigg = 28 . 4146 . With the new speed v 0 and angle θ , the new range x can be found similarly to the previous part as x = v 2 o sin 2 θ 2 g ± radicalBig v 4 o sin 2 2 θ + 8 g v 2 o h cos 2 θ 2 g . Since v 4 o sin 2 2 θ + 8 g v 2 o h cos 2 θ = (95 . 237 m / s) 4 (sin 2 28 . 4146 ) +(8) (9 . 81 m / s 2 ) (95 . 237 m / s) 2 × (2 . 5 m) (cos 2 28 . 4146 ) = 5 . 90159 × 10 7 m 2 / s 2 , then x = (95 . 237 m / s) 2 (sin 28 . 4146 ) 2 (9 . 81 m / s 2 ) + radicalbig 5 . 90159 × 10 7 m 2 / s 2 2 (9 . 81 m / s 2 ) = 778 . 503 m .
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