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Unformatted text preview: fullwood (tf5349) Homework #4 Projectile Catala (49033) 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A fireman, 56 . 9 m away from a burning build ing, directs a stream jet of water from a ground level fire hose at an angle of 30 . 8 above the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . If the speed of the stream as it leaves the hose is 40 . 2 m / s, at what height will the stream of water strike the building? Correct answer: 20 . 614 m. Explanation: 56 . 9 m 4 . 2 m / s 30 . 8 y Given : d = 56 . 9 m , v i = 40 . 2 m / s , and = 30 . 8 . The components of the initial velocity are v ix = v i cos , and v iy = v i sin . The horizontal motion carries the water to the building, so x = v ix t = ( v i cos ) t t = x v i cos = 56 . 9 m (40 . 2 m / s)(cos 30 . 8 ) = 1 . 64783 s . The height the water reaches at this time is y = v iy t 1 2 g t 2 = ( v i sin ) t 1 2 g t 2 = (40 . 2 m / s)(sin30 . 8 ) (1 . 64783 s) 1 2 ( 9 . 8 m / s 2 ) (1 . 64783 s) 2 = 20 . 614 m . 002 10.0 points A daredevil decides to jump a canyon. Its walls are equally high and 10 m apart. He takes off by driving a motorcycle up a short ramp sloped at an angle of 14 . The acceleration of gravity is 9 . 8 m / s 2 . What minimum speed must he have in or der to clear the canyon? Correct answer: 14 . 448 m / s. Explanation: 14 10 m Note: Figure is not drawn to scale Given : x = 10 m , = 14 , and g = 9 . 8 m / s 2 . We can find the time of flight from the hori zontal motion: x = v i (cos ) t t = x v i (cos ) . From the vertical motion, y = v i (sin ) t 1 2 g ( t ) 2 = 0 v i (sin ) 1 2 g t = 0 fullwood (tf5349) Homework #4 Projectile Catala (49033) 2 so that v i (sin ) = 1 2 g bracketleftbigg x v i (cos ) bracketrightbigg v 2 i = g x 2 (sin ) (cos ) v i = radicalBigg g x 2 (sin ) (cos ) = radicalBigg (9 . 8 m / s 2 ) (10 m) 2 (cos14 ) (sin14 ) = 14 . 448 m / s . 003 10.0 points Cliff divers at Acapulco jump into the sea from a cliff 37 . 4 m high. At the level of the sea, a rock sticks out a horizontal distance of 14 . 79 m. The acceleration of gravity is 9 . 8 m / s 2 . With what minimum horizontal velocity must the cliff divers leave the top of the cliff if they are to miss the rock? Correct answer: 5 . 35341 m / s. Explanation: Given : h = 37 . 4 m and x = 14 . 79 m . The time of flight is defined by the vertical drop: h = v iy t 1 2 g t 2 = 1 2 g t 2 since v iy = 0 m/s, so t = radicalBigg 2 h g The horizontal velocity is constant, so x = v x t v x = x t = x radicalbigg g 2 h = 14 . 79 m radicalBigg 9 . 8 m / s 2 2 ( 37 . 4 m) = 5 . 35341 m / s ....
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