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2D Kinematics II-solutions

# 2D Kinematics II-solutions - le(txl65 2D Kinematics II...

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le (txl65) – 2D Kinematics II – mohanty – (81440) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A plane drops a hamper of medical supplies from a height of 2690 m during a practice run over the ocean. The plane’s horizontal veloc- ity was 131 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: 264 . 358 m / s. Explanation: BasicConcept: 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 = 131 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 ) (2690 m) = 229 . 617 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 (229 . 617 m / s ) 2 + (131 m / s ) 2 = 264 . 358 m / s 002 10.0points A toy cannon fires a 0 . 099 kg shell with initial velocity v i = 8 . 8 m / s in the direction θ = 46 above the horizontal. Δ x Δ h 8 . 8 m / s 46 Δ y y The shell’s trajectory curves downward be- cause of gravity, so at the time t = 0 . 549 s the shell is below the straight line by some vertical distance Δ h . Find this distance Δ h in the absence of air resistance. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 47686 m. Explanation: In the absence of gravity, the shell would fly along the straight line at constant velocity: ˆ x = t v i cos θ , ˆ y = t v i sin θ . The gravity does not affect the x coordinate of the shell, but it does pull its y coordinate at constant downward acceleration a y = - g , hence x = t v i cos θ, y = t v i sin θ - g t 2 2 . Thus, x = ˆ x but y = ˆ y - 1 2 gt 2 , or in other words, the shell deviates from the straight-line path by the vertical distance Δ h = | ˆ y - y | = g t 2 2 . Note: This result is completely indepen- dent on the initial velocity v i or angle θ of the shell. It is a simple function of the flight time t and nothing else (besides the constant g = 9 . 8 m / s 2 ).

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le (txl65) – 2D Kinematics II – mohanty – (81440) 2 Δ h = g t 2 2 = (9 . 8 m / s 2 ) (0 . 549 s) 2 2 = 1 . 47686 m .
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2D Kinematics II-solutions - le(txl65 2D Kinematics II...

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