Therefore, its elicit is never vertical and is never parallel to the acceleration. (b) Yes. A projectile at the top of its trajectory has a velocity that is horizontal, while at the same time its acceleration is vertical.

6. Just before it lands, this projectile is moving downward with the same speed it had when it was launched. In addition, if it was launched upward at an angle e above the x axis, it is moving in a direction that is an angle 0 below the x axis just before it lands. Therefore, its velocity just before landing is v ( 2 ran/s ) x” + (-mm/s)y- 8.

- Thesis Statement
- Structure and Outline
- Voice and Grammar
- Conclusion

Maximum height depends on the initial speed squared. Therefore, to reach twice the height, projectile 1 must have an initial speed that is the square root of 2 times greater than the initial speed of projectile 2. It follows that the ratio of the speeds is the square root of 2. 10. The tomato lands on the road in front of you.

This follows from the fact that its horizontal speed is the same as yours during the entire time of its fall. Solutions to Problems and Conceptual Exercises Picture the Problem: You walk briskly down the street while tossing a ball in the air and catching it again.Strategy: Use a separate analysis of the irrational and vertical motions of the ball to answer the conceptual question. Solution: 1 . (a) As long as air friction is neglected there is no acceleration of the ball in the horizontal direction. The ball will continue moving horizontally with the same speed as your walking speed. Therefore, you need to launch the ball straight upward relative to yourself in order for it to land back in your hand.

2. (b) The best explanation (see above) is Ill. The ball moves in the forward direction with your walking speed at all times.Statements and II are false because they ignore the inertia of the ball in the horizontal direction.

Insight: [f air friction is taken into account you must launch the ball in the forward direction a little bit. While it is in the air the friction will slow the ball horizontally so that it lands back in your hand. Picture the Problem: The vector involved in this problem is depicted at right. Strategy: Separate v into x- and y-components.

Let north be along the y-axis, west along the -?x axis. Find the components of the velocity in each direction and use them to find the distances traveled.Solution: 1. (a) Find vs. and v y ; vs.=-? (4. 2 m/s ) coos 320 = -3.

56 m/s v Y -? ( 4. 2 m/s ) sin 320 2. 3 m/s 2. Find the westward distance traveled: y = 3. 56 m/s 25 min x 60 s/min ) 3. (b) Find the northward x=vet=( 2.

23 m/s )(25 min x 60 s/min ) ivy w 320 = 5300 m = 5. 3 km = 3300 m = 3. 3 km Insight: The northward and westward motions can be considered separately. In this case they are each described by constant velocity motion. Copyright 0 2010 Pearson Education, Inc. All rights reserved.

This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. James S. Walker, Physics, 4th Edition 3. Picture the Problem: The vectors involved in this problem are depicted at right. Strategy: Let north be along the y-axis and east along the x-axis. Find the components of the velocity in each direction, and use them to find the times elapsed.

Solution: 1. (a) Find the x component of v : vs. = (1. 75 m/s ) cosec. 00 = 1. 66 18. 00 20. 0 m = 12.

05 vs. 1. 66 m/s 2. Find the time elapsed to travel east 20. 0 m: 3. B) Find they component of v: -0. 541 m/s 4. Find the time elapsed to travel north 30.

0 m: 30. 0 m 55. 5 s ivy. 541 or,/s Insight: The northward and eastward motions can be considered separately. In both cases the actual distance traveled is greater than 20.

0 or 30. 0 m, respectively. For instance, in the second case you must actually travel a total distance Of 30. 0 m sin 18. 00 97.

1 m to change your displacement by 30. 0 m north. Picture the Problem: The car moves up the 5. 50 incline with constant acceleration, changing both its horizontal and vertical displacement simultaneously.Strategy: Find the magnitude of the displacement along the incline, and then independently find the horizontal and vertical components of the escapement.

Solution: 1. (a) Find the magnitude of the displacement along the incline using equation 2-11: 2. Find the horizontal component of Ar : coos (140 m ) coos 5.

50 = m 3. (b) Find the vertical component Ar : y = d sin e m) sin 5. 50 = mom 2 = mom Insight: The horizontal and vertical motions can be considered separately. In this case they are each described by constant acceleration motion, but the vertical acceleration is less than the horizontal.The two accelerations would be equal if the angle of the incline were 450. 5. Picture the Problem: The motion of the particle is depicted at right. Strategy: use the given information to independently write the equations of motion in the x and y directions.

There will be a pair of equations for the position of the particle (like equation 4-6) and a pair for the velocity (like equation 4-6), except in this case the acceleration will not be the same as g . Solution: 1. (a) Write equation 4-6 for the x position of the particle using a = (-4. 4 m/s 2) x* instead of g: x=xx+box+exact 2.Now do the same for the y coordinate of the position: ran,’s 3. (b) Write equation 4-6 for the x component of the velocity using a = ( -? 4.

4 m/s 2 ) x- instead of g: + 2) (5. 0 s) = -22 ms 4. Now do the same for the y component of the velocity: 6. 2 12 (-?4. 4 2 x = -55 m +0 = 6. 2 ms 5. (c) Because vs. continually increases in the negative direction, the speed will increase with time.

Insight: This problem is very much akin to projectile motion, with uniform acceleration in one direction and constant velocity the perpendicular direction. The only difference is the acceleration is – 4. 4 m/s 2 and is in the x direction.Copyright 0 201 0 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Picture the Problem: The motion of the electron is depicted at right. Position of the particle (like equation 4-6), except the acceleration will not be the same as g . Use the equations of motion to find the requested time and position information. Let the x direction correspond to horizontal, and the y direction to vertical. 6. 20 -O = 2.

95 x -95 2. 0 x 109 CM/s Solution: 1 . (a) The horizontal motion is characterized by constant velocity. Apply equation 4-6: . (b) use the time to find the vertical deflection , again using equation 4-6 except substituting a for g. Y=you+vote+12th = O CM +0 CM+ 12 (5. 30 x 1017 10-9 s) 2.

31 CM Insight: This problem is very much akin to projectile motion, with uniform acceleration in one direction and constant velocity the perpendicular direction. The only difference is the acceleration is 5. Xx 1017 CM/so upward instead of 9. 81 m/so downward.

7. Picture the Problem: The paths of the two canoeists are shown at right.Strategy: Canoeist Xi’s 450 path determines an isosceles right triangle whose eggs measure 1. 0 km. So canoeist g’s path determines a right triangle whose legs measure 1.

0 km north and 0. 5 km west. Use the right triangle to find the angle B. Find the distance traveled by each canoeist and set the times of travel equal to each other to determine the appropriate speed of canoeist 2: CLC 0. 5 kronor 270 Solution: 1. (a) Find the angle e from the right triangle of canoeist 2: e -?tan-1 0 2. (b) Set the travel times equal to each other: 3. Use the resulting ratio to find the appropriate speed of canoeist 2: Earl Are v d v=ova ( 0.

Km ) + (1. 0 km ) (1. 0 km) 01. 5 0 = 1.

1 m/s Insight: There are other ways to solve this problem. For instance, because the motions are independent, we could set the time it takes canoeist 1 to travel 1. 0 km horizontally equal to the time it takes canoeist 2 to travel 0. 5 km horizontally. Picture the Problem: Two divers run horizontally off the edge of a low cliff. Strategy: Use a separate analysis of the horizontal and vertical motions of the divers to answer the conceptual question. Solution: 1. (a) As long as air friction is neglected there is no acceleration of either diver in the horizontal direction.

The divers will continue moving horizontally at the same speed with which they left the cliff. However, the time of flight for each diver will be identical because they fall the same vertical distance. Therefore, diver 2 will travel twice as much horizontal distance as diver 1 2. (b) The best explanation (see above) is I.

The drop time is the same for both divers. Statement II is true but not relevant. Statement Ill is false because the total distance covered depends upon the horizontal speed. Insight: If air friction is taken into account diver 2 will travel less than twice the horizontal distance as diver 1 .This is because air friction is proportional to speed, so diver 2, traveling at a higher speed, will experience a larger force. 4-3 9.

Picture the Problem: Two diversified off an overhang into a lake. Diver 1 drops straight down, and diver 2 runs off the cliff with an initial horizontal speed vow Strategy: use a separate analysis of the horizontal and vertical motions of the identical because they fall the same vertical distance. Therefore, the splashdown speed of diver 2 will be greater than the splashdown speed of diver 1. 2.

(b) The best explanation (see above) is II.