HW02 solutions

# HW02 solutions - Kwok (mk2749) – HW02 – Tsoi –...

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Unformatted text preview: Kwok (mk2749) – HW02 – Tsoi – (58020) 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 (part 1 of 2) 10.0 points The cartesian coordinates of a point in the xy plane are x = − 6 . 57 m, y = − 2 . 9 m. Find the distance, r , from the point to the origin. Correct answer: 7 . 18157 m. Explanation: To find the distance from the point to the origin, we use the formula r = radicalbig x 2 + y 2 , which gives us r = radicalBig ( − 6 . 57 m ) 2 + ( − 2 . 9 m ) 2 = 7 . 18157 m . 002 (part 2 of 2) 10.0 points Calculate the angle θ between the radius- vector of the point and the positive x axis (measured counterclockwise from the positive x axis, within the limits of − 180 ◦ to +180 ◦ ). Correct answer: − 156 . 183 ◦ . Explanation: The formula tan θ = y x and the fact that the point is in the third quadrant of the coordinate system (since the cartesian coordinates x and y of the point are negative), yield θ = arctan y x − 180 ◦ = arctan ( − 2 . 9 m ) ( − 6 . 57 m ) − 180 ◦ = − 156 . 183 ◦ . That is, θ = − 156 . 183 ◦ . 003 10.0 points A car is parked near a cliff overlooking the ocean on an incline that makes an angle of 22 . 7 ◦ with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has a velocity 20 m / s when it reaches the edge of the cliff. The cliff is 42 . 6 m above the ocean. The acceleration of gravity is 9 . 8 m / s 2 . θ How far is the car from the base of the cliff when the car hits the ocean? Correct answer: 41 . 7788 m. Explanation: First, find the car’s initial vertical velocity when it leaves the cliff v y = v sin θ = (20 m / s) sin(22 . 7 ◦ ) = 7 . 71812 m / s . Then find the vertical velocity with which the car strikes the water as v 2 y = v 2 y + 2 g h = (7 . 71812 m / s) 2 + 2 (9 . 8 m / s 2 ) (42 . 6 m) = 894 . 529 m 2 / s 2 , or v y = 29 . 9087 m / s . The time of flight is found from v y = v y + g t as t = v y − v y g = 2 . 26434 s . The initial velocity is v x = v cos θ and the horizontal motion of the car during this time is x = v x t = (20 m / s) cos(22 . 7 ◦ ) (2 . 26434 s) = 41 . 7788 m . Kwok (mk2749) – HW02 – Tsoi – (58020) 2 004 10.0 points A target lies flat on the ground 10 m from the side of a building that is 10 m tall, as shown below. The acceleration of gravity is 10 m / s 2 . Air resistance is negligible. A student rolls a 10 kg ball off the horizontal roof of the building in the direction of the target. 10 m 10m v 10m The horizontal speed v with which the ball must leave the roof if it is to strike the target is most nearly 1. v = √ 5 10 m/s....
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## This note was uploaded on 02/10/2009 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas.

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HW02 solutions - Kwok (mk2749) – HW02 – Tsoi –...

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