This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: le (txl65) 2D Kinematics II mohanty (81440) 1 This printout should have 15 questions. Multiplechoice 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 2690 m during a practice run over the ocean. The planes 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: 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 = 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 10.0 points A toy cannon fires a 0 . 099 kg shell with initial velocity v i = 8 . 8 m / s in the direction = 46 above the horizontal. x h 8 . 8 m / s 4 6 y y The shells 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 h . Find this distance 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: x = t v i cos , y = t v i sin . 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 , y = t v i sin  g t 2 2 . Thus, x = x but y = y 1 2 gt 2 , or in other words, the shell deviates from the straightline path by the vertical distance h =  y y  = g t 2 2 . Note: This result is completely indepen dent on the initial velocity v i or angle 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 ). le (txl65) 2D Kinematics II mohanty (81440) 2 h = g t 2 2 = (9 . 8 m / s 2 ) (0 . 549 s) 2 2 =...
View
Full
Document
This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.
 Spring '11
 sdsd

Click to edit the document details