# HW5 - schneiter (tcs639) – HW05 – Ross – (89152) 1...

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Unformatted text preview: schneiter (tcs639) – HW05 – Ross – (89152) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A plane drops a hamper of medical supplies from a height of 3460 m during a practice run over the ocean. The plane’s horizontal veloc- ity was 141 m / s at the instant the hamper was dropped. The acceleration of gravity is 9 . 8 m / s 2 . What is the magnitude of the overall veloc- ity of the hamper at the instant it strikes the surface of the ocean? Correct answer: 296 . 137 m / s. Explanation: Basic Concept: Motion in gravity field v 2 = 2 g h Solution: This is a projectile motion problem. The motion of the dropping hamper consists of two parts: horizontally, it moves with the initial velocity of the plane, i.e. v h = v = 141 m / s ; vertically, due to Gravity, it moves as a freely falling body. Applying the equation above gives the vertical velocity as v v = radicalbig 2 g h = radicalBig 2 (9 . 8 m / s 2 ) (3460 m) = 260 . 415 m / s Thus the overall velocity at the instant the hamper strikes the surface of the ocean is v f = radicalBig v 2 v + v 2 h = radicalBig (260 . 415 m / s ) 2 + (141 m / s ) 2 = 296 . 137 m / s 002 (part 1 of 2) 10.0 points The speed of an arrow fired from a compound bow is about 29 m / s. An archer sits astride his horse and launches an arrow into the air, elevating the bow at an angle of 35 ◦ above the horizontal and 2 . 6 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 2 9 m / s 3 5 ◦ 2 . 6 m height range What is the arrow’s range? Assume: The ground is level. Ignore air resistance. Correct answer: 84 . 115 m. Explanation: Let : v o = 29 m / s , θ = 35 ◦ , h o = 2 . 6 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 o h x In the projectile motion, we have x = ( v o cos θ ) t t = x v o cos θ horizontally, and y = h o + ( v o sin θ ) t- 1 2 g t 2 vertically. Thus y = h o + x v o sin θ v o cos θ- 1 2 g x 2 v 2 o cos 2 θ = 0...
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## This note was uploaded on 04/23/2009 for the course PHYS 152 taught by Professor Button during the Spring '08 term at IUPUI.

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HW5 - schneiter (tcs639) – HW05 – Ross – (89152) 1...

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