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05 kinematics in two dimension
- 1. Kinematics in Two Dimension Topic 3
- 3. Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.
- 5. It can be understood by analyzing the horizontal and vertical motions separately.
- 6. The speed in the x -direction is constant; in the y -direction the object moves with constant acceleration g . This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x -direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.
- 7. If an object is launched at an initial angle of θ 0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.
- 8. Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is down.
- 9. Example 3-6: Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
- 12. Example 3-11: A punt. Suppose the football in Example 3–7 was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground? Set x 0 = 0, y 0 = 0.
- 15. Here, v WS is the velocity of the water in the shore frame, v BS is the velocity of the boat in the shore frame, and v BW is the velocity of the boat in the water frame. The relationship between the three velocities is:
- 17. Example 3-14: Heading upstream. A boat’s speed in still water is v BW = 1.85 m/s. If the boat is to travel directly across a river whose current has speed v WS = 1.20 m/s, at what upstream angle must the boat head?
- 19. Example 3-16: Car velocities at 90 °. Two automobiles approach a street corner at right angles to each other with the same speed of 40.0 km/h (= 11.1 m/s), as shown. What is the relative velocity of one car with respect to the other? That is, determine the velocity of car 1 as seen by car 2.
Hinweis der Redaktion
- Figure 3-19. Caption: This strobe photograph of a ball making a series of bounces shows the characteristic “parabolic” path of projectile motion.
- Figure 3-20. Caption: Projectile motion of a small ball projected horizontally. The dashed black line represents the path of the object. The velocity vector at each point is in the direction of motion and thus is tangent to the path. The velocity vectors are green arrows, and velocity components are dashed. (A vertically falling object starting at the same point is shown at the left for comparison; v y is the same for the falling object and the projectile.)
- Figure 3-21. Caption: Multiple-exposure photograph showing positions of two balls at equal time intervals. One ball was dropped from rest at the same time the other was projected horizontally outward. The vertical position of each ball is seen to be the same at each instant.
- Figure 3-22. Caption: Path of a projectile fired with initial velocity v 0 at angle θ 0 to the horizontal. Path is shown dashed in black, the velocity vectors are green arrows, and velocity components are dashed. The acceleration a = dv/dt is downward. That is, a = g = -gj where j is the unit vector in the positive y direction.
- Figure 3-23. Answer: The x velocity is constant; the y acceleration is constant. We know x 0 , y 0 , x, y, a, and v y0 , but not v x0 or t. The problem asks for v x0 , which is 28.2 m/s.
- Figure 3-24. Answers: a. This can be calculated using vertical variables only, giving 7.35 m. b. The equation for t is quadratic, but the only meaningful solution is 2.45 s (ignoring the t = 0 solution). c. The x velocity is constant, so x = 39.2 m. d. At the highest point, the velocity is the (constant) x velocity, 16.0 m/s. e. The acceleration is constant, 9.80 m/s 2 downward.
- Figure 3-27. Caption: Example 3–10. (a) The range R of a projectile; (b) there are generally two angles θ 0 that will give the same range. Can you show that if one angle is θ 01 , the other is θ 0 2 = 90° - θ 0 1 ? Answer: a. Eliminating t from the kinematics equations, and solving for x when y = 0, we get R = (v 0 2 sin 2 θ 0 ) /g. b. There is a unique solution for sin 2 θ 0 , but two solutions for the angle: θ 0 = 30.3 ° or 59.7 °.
- Figure 3-28. Caption: Example 3–11: the football leaves the punter’s foot at y = 0, and reaches the ground where y = –1.00 m. Answer: We can find t from the vertical motion; ignoring the negative solution in the quadratic gives t = 2.53 s. Then we find the horizontal distance to be 40.5 m.
- Figure 3-29. Answer: a. We can find the time the package will take to fall, and then the horizontal distance it will travel: 450 m b. We know the horizontal distance and the horizontal speed (the speed of the helicopter), so we can find the time the package will take to fall. Substituting gives the initial y velocity: -7.0 m/s (downward) c. 94 m/s
- Figure 3-31. Caption: To move directly across the river, the boat must head upstream at an angle θ . Velocity vectors are shown as green arrows: v BS = velocity of B oat with respect to the S hore, v BW = velocity of B oat with respect to the W ater, v WS = velocity of the W ater with respect to the S hore (river current).
- Figure 3-33. Answer: 40.4 ° (simple geometry)
- Figure 3-34. Caption: Example 3–15. A boat heading directly across a river whose current moves at 1.20 m/s. Answer: a. 2.21 m/s at an angle of 33.0 ° downstream. b. 59.5 s and 71.4 m
- Figure 3-35. Answer: 15.7 m/s