PHY303k_-_Spring_2010_-_HW_5

PHY303k_-_Spring_2010_-_HW_5 - saldana (avs387) Homework 05...

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Unformatted text preview: saldana (avs387) 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 18 m / s. An archer sits astride his horse and launches an arrow into the air, elevating the bow at an angle of 67 above the horizontal and 1 . 5 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 1 8 m / s 6 7 1 . 5 m range What is the arrows range? Assume: The ground is level. Ignore air resistance. Correct answer: 24 . 3785 m. Explanation: Let : v o = 18 m / s , = 67 , h = 1 . 5 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 h range In the projectile motion, we have x = ( v o cos ) t t = x v o cos horizontally, and y = h + ( v o sin ) t- 1 2 g t 2 vertically. Thus y = h + x v o sin v o cos - 1 2 g x 2 v 2 o cos 2 = 0 when the arrow lands. Thus 2 v 2 o h cos 2 + 2 x v 2 o sin cos - g x 2 = 0- 2 v 2 o h cos 2 - x v 2 o sin2 + g x 2 = 0 x = v 2 o sin 2 radicalBig v 4 o sin 2 2 + 8 g v 2 o h cos 2 2 g . Since v 4 o sin 2 2 + 8 g v 2 o h cos 2 = (18 m / s) 4 (sin 2 134 ) + (8) (9 . 81 m / s 2 ) (18 m / s) 2 (1 . 5 m) (cos 2 67 ) = 60142 . 9 m 2 / s 2 , then x = (18 m / s) 2 (sin134 ) 2 (9 . 81 m / s 2 ) + radicalbig 60142 . 9 m 2 / s 2 2 (9 . 81 m / s 2 ) = 24 . 3785 m . saldana (avs387) Homework 05 florin (58140) 2 002 (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 horses speed is 21 m / s? Correct answer: 97 . 1692 m. Explanation: The arrows speed relative to the ground and the angle of elevation relative to the ground are changed v x = v arrow + v archer = (18 m / s) cos 67 + 21 m / s = 28 . 0332 m / s . v y = (18 m / s) sin 67 = 16 . 5691 m / s The arrows speed relative to the ground is then given by v o = radicalBig v 2 x + v 2 y = radicalBig (28 . 0332 m / s) 2 + (16 . 5691 m / s) 2 = 32 . 5637 m / s , and the angle of elevation relative to the ground is = tan 1 v y v x = tan 1 parenleftbigg 16 . 5691 m / s 28 . 0332 m / s parenrightbigg = 30 . 5853 . With the new speed v and angle , the new range x can be found similarly to the previous part as x = v 2 o sin 2 2 g radicalBig v 4 o sin 2 2 + 8 g v 2 o h cos 2 2 g ....
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This note was uploaded on 08/30/2010 for the course PHY 58140 taught by Professor Florin during the Spring '10 term at University of Texas at Austin.

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PHY303k_-_Spring_2010_-_HW_5 - saldana (avs387) Homework 05...

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