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Unformatted text preview: banuelos (ojb93) – Homework 05 – florin – (58140) 1 This printout should have 13 questions. Multiplechoice 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 55 m / s. An archer sits astride his horse and launches an arrow into the air, elevating the bow at an angle of 60 ◦ above the horizontal and 1 . 1 m above the ground. The acceleration of gravity is 9 . 81 m / s 2 . b b b b b b b b b b b b b b b b b b b b 5 5 m / s 6 ◦ 1 . 1 m range What is the arrow’s range? Assume: The ground is level. Ignore air resistance. Correct answer: 267 . 68 m. Explanation: Let : v o = 55 m / s , θ = 60 ◦ , h = 1 . 1 m , and g = 9 . 81 m / s 2 . b b b b b b b b b b b b b b b b b b b b 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 sin2 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 = (55 m / s) 4 (sin 2 120 ◦ ) + (8) (9 . 81 m / s 2 ) × (55 m / s) 2 (1 . 1 m) (cos 2 60 ◦ ) = 6 . 92825 × 10 6 m 2 / s 2 , then x = (55 m / s) 2 (sin120 ◦ ) 2 (9 . 81 m / s 2 ) + radicalbig 6 . 92825 × 10 6 m 2 / s 2 2 (9 . 81 m / s 2 ) = 267 . 68 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 15 m / s? Correct answer: 413 . 688 m. Explanation: banuelos (ojb93) – Homework 05 – florin – (58140) 2 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 = (55 m / s) cos 60 ◦ + 15 m / s = 42 . 5 m / s . v y = (55 m / s) sin 60 ◦ = 47 . 6314 m / s The arrow’s speed relative to the ground is then given by v ′ o = radicalBig v 2 x + v 2 y = radicalBig (42 . 5 m / s) 2 + (47 . 6314 m / s) 2 = 63 . 8357 m / s , and the angle of elevation relative to the ground is θ ′ = tan − 1 v y v x = tan − 1 parenleftbigg 47 . 6314 m / s 42 . 5 m / s parenrightbigg = 48 . 2585 ◦ . With the new speed v ′ 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 ....
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This note was uploaded on 04/11/2010 for the course PHYSICS 58100 taught by Professor Florin during the Spring '10 term at University of Texas.
 Spring '10
 Florin
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