Practice 02 Solutions

# Practice 02 Solutions - practice 02 – GADHIA TEJAS –...

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Unformatted text preview: practice 02 – GADHIA, TEJAS – Due: Jan 24 2007, 5:00 pm 1 Question 1 part 1 of 2 10 points A bullet is fired straight down from the top of a high cliff. What is the acceleration of the bullet? 1. First faster than 9.8 m/s 2 , then slower. 2. First slower than 9.8 m/s 2 , then faster. 3. Equal to 9.8 m/s 2 . correct 4. Less than 9.8 m/s 2 . 5. Not enough information to answer. 6. More than 9.8 m/s 2 . Explanation: Regardless of whether the bullet is dropped, fired straight up, or fired straight down, it is under the influence of gravitational accelera- tion g = 9 . 8 m / s 2 . Question 2 part 2 of 2 10 points What conclusion can be reached about the speed of the bullet? 1. Not enough information to answer. 2. Always faster than its initial speed. cor- rect 3. Equal to zero. 4. Equal to its initial speed. 5. First faster than its initial speed, then slower. 6. Equal to 9.8 m/s 2 . 7. Always slower than its initial speed. 8. First slower than its initial speed, then faster. 9. Equal to 9.8 m/s. Explanation: Gravity will continue to accelerate it faster than its initial speed. Question 3 part 1 of 1 10 points A ball tossed vertically upward rises, reaches its highest point, and then falls back to its starting point. During this time the acceleration of the ball is 1. directed upward initially, then directed downward. 2. in the direction of motion. 3. opposite its velocity. 4. directed downward. correct 5. directed upward. Explanation: The acceleration of an object due to gravity is always directed toward the earth’s surface. Question 4 part 1 of 1 10 points A bullet is fired straight up from a gun with a muzzle velocity of 230 m / s. The acceleration of gravity is 9 . 8 m / s 2 . Neglecting air resistance, what will be its displacement after 4 s? Correct answer: 841 . 6 m (tolerance ± 1 %). Explanation: Displacement is defined by s = s o + v o t + 1 2 a t 2 . The initial velocity of the bullet is v o = v and its initial displacement is s o = 0 . The gravitational acceleration is negative, so the displacement is s = s o + v o t − 1 2 g t 2 practice 02 – GADHIA, TEJAS – Due: Jan 24 2007, 5:00 pm 2 = v t − 1 2 g t 2 = (230 m / s) (4 s) − 1 2 g (4 s) 2 = 841 . 6 m . Question 5 part 1 of 3 10 points A hiker begins a trip by first walking 15 . 2 km southeast from her base camp. On the second day, she walks 46 . 1 km in a direc- tion 71 . 4 ◦ north of east, at which point she discovers a forest ranger’s tower. Determine the y component of the total displacement. Correct answer: 32 . 9441 km (tolerance ± 1 %). Explanation: Let us denote the displacement vectors on the first and second days by A and B , respec- tively, and use the camp as the origin of the coordinates. Let the angle between A and the positive x direction be α , and the angle between B and the positive x direction be β ....
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## This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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Practice 02 Solutions - practice 02 – GADHIA TEJAS –...

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