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# solution27 - fullwood(tf5349 Homework#4 Projectile...

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fullwood (tf5349) – Homework #4 Projectile – Catala – (49033) 1 This print-out should have 22 questions. Multiple-choice 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 40 . 2 m / s 30 . 8 Δ y Given : d = 56 . 9 m , v i = 40 . 2 m / s , and θ = 30 . 8 . The components of the initial velocity are v ix = v i cos θ , and v iy = v i sin θ . The horizontal motion carries the water to the building, so Δ x = v ix t = ( v i cos θ ) t t = Δ x v i cos θ = 56 . 9 m (40 . 2 m / s)(cos 30 . 8 ) = 1 . 64783 s . The height the water reaches at this time is Δ y = v iy t 1 2 g t 2 = ( v i sin θ ) t 1 2 g t 2 = (40 . 2 m / s)(sin 30 . 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 : Δ x = 10 m , θ = 14 , and g = 9 . 8 m / s 2 . We can find the time of flight from the hori- zontal motion: Δ x = v i (cos θ ) Δ t Δ t = Δ x v i (cos θ ) . From the vertical motion, Δ y = v i (sin θ ) Δ t 1 2 g t ) 2 = 0 v i (sin θ ) 1 2 g Δ t = 0

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fullwood (tf5349) – Homework #4 Projectile – Catala – (49033) 2 so that v i (sin θ ) = 1 2 g bracketleftbigg Δ x v i (cos θ ) bracketrightbigg v 2 i = g Δ x 2 (sin θ ) (cos θ ) v i = radicalBigg g Δ x 2 (sin θ ) (cos θ ) = radicalBigg (9 . 8 m / s 2 ) (10 m) 2 (cos 14 ) (sin 14 ) = 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 : Δ h = 37 . 4 m and Δ 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 Δ x = v x Δ t v x = Δ x Δ t = Δ x radicalbigg g 2 Δ h = 14 . 79 m radicalBigg 9 . 8 m / s 2 2 ( 37 . 4 m) = 5 . 35341 m / s . 004 10.0 points A ball is thrown and follows the parabolic path shown. Air friction is negligible. Point Q is the highest point on the path. Points P and R are the same height above the ground. Q R P Which of the following diagrams best indi- cates the direction of the acceleration, if any, on the ball at point P ?
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solution27 - fullwood(tf5349 Homework#4 Projectile...

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