QTest #3-solutions

# QTest #3-solutions - rooF oF the building in the direction...

This preview shows pages 1–2. Sign up to view the full content.

Version 140/ACADA – QTest #3 – Antoniewicz – (57380) 1 This print-out should have 2 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A man can throw a ball a maximum horizontal distance oF 146 m. The acceleration oF gravity is 9 . 8 m / s 2 . How Far can he throw the same ball verti- cally upward with the same initial speed? 1. 69.0 2. 83.0 3. 64.0 4. 68.0 5. 95.0 6. 73.0 7. 54.5 8. 49.05 9. 89.0 10. 76.0 Correct answer: 73 m. Explanation: The range oF a particle is given by the ex- pression R = v 2 0 sin 2 θ g . The maximum horizontal distance is obtained when the ball is thrown at an angle θ = 45 and sin 2 θ = 1 . Solving For v 0 , v 0 = r g R . When the ball is thrown upward with this speed, the maximum height is obtained From the equation v 2 f = v 2 0 - 2 g h . Let v f = 0, and solve For h h = v 2 0 2 g . 002 10.0 points A target lies ±at on the ground 5 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 5 kg ball o² the horizontal

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rooF oF the building in the direction oF the target. 5 m 10 m v 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 2 m/s. 2. v = 5 2 2 m/s. correct 3. v = 5 5 m/s. 4. v = 3 5 m/s. 5. v = 5 3 m/s. 6. v = 5 3 3 m/s. 7. v = 2 5 m/s. 8. v = 2 5 m/s. 9. v = 3 5 m/s. 10. v = 5 5 m/s. Explanation: m = 5 kg , not required h = 10 m , x = 5 m , and g = 10 m / s 2 . Version 140/ACADA QTest #3 Antoniewicz (57380) 2 Observe the motion in the vertical direction only and it is a purely 1-dimension movement with a constant acceleration. So the time need for the ball to hit the ground is t = r 2 h g and the horizontal speed should be v = x t for the ball to hit the target. Therefore v = x R g 2 h = (5 m) r 10 m / s 2 2 (10 m) = 5 2 m / s = 5 2 2 m / s ....
View Full Document

## This note was uploaded on 03/01/2011 for the course PHYS 317K taught by Professor Turner during the Spring '11 term at University of Texas at Austin.

### Page1 / 2

QTest #3-solutions - rooF oF the building in the direction...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online