05 kinematics in two dimension
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This document outlines key concepts in two-dimensional kinematics including projectile motion, relative velocity, and provides examples of applying these concepts. It discusses how projectile motion can be analyzed by separating the horizontal and vertical components of motion. Relative velocity is defined as the velocity of an object with respect to a reference frame. Several examples are given that apply concepts of projectile motion and relative velocity to calculate values like maximum height, time of travel, range, and relative velocities between objects.
05 kinematics in two dimension
- 1. Kinematics in Two Dimension Topic 3
- 2. Lecture Outline Projectile Motion Relative Velocity
- 3. Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.
- 4. Galileo Galilei (1564 – 1642) was the first to describe projectile motion Could be understood by analyzing the horizontal and vertical component of the motion separately
- 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.
- 10. Example 3-7: A kicked football. A football is kicked at an angle θ 0 = 37.0° with a velocity of 20.0 m/s, as shown. Calculate the maximum height, the time of travel before the football hits the ground, how far away it hits the ground, the velocity vector at the maximum height, and the acceleration vector at maximum height. Assume the ball leaves the foot at ground level, and ignore air resistance and rotation of the ball.
- 11. Example 3-10: Level horizontal range. Derive a formula for the horizontal range R of a projectile in terms of its initial speed v 0 and angle θ 0 . The horizontal range is defined as the horizontal distance the projectile travels before returning to its original height (which is typically the ground); that is, y (final) = y 0 . Suppose one of Napoleon’s cannons had a muzzle speed, v 0 , of 60.0 m/s. At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?
- 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.
- 13. Example 3-12: Rescue helicopter drops supplies. A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), how far in advance of the recipients (horizontal distance) must the package be dropped? Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? With what speed does the package land in the latter case?
- 14. Relative Velocity Relative velocity is the velocity of an object with respect to a certain reference frame
- 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:
- 16. For any two object, or reference frame, A and B, the velocity A relative to B has the same magnitude, but opposite direction, as the velocity of B relative to A. V BA = - V AB
- 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?
- 18. Example 3-15: Heading across the river. The same boat (v BW = 1.85 m/s) now heads directly across the river whose current is still 1.20 m/s. What is the velocity (magnitude and direction) of the boat relative to the shore? If the river is 110 m wide, how long will it take to cross and how far downstream will the boat be then?
- 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.
Editor's Notes
- #4: Figure 3-19. Caption: This strobe photograph of a ball making a series of bounces shows the characteristic “parabolic” path of projectile motion.
- #6: 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.)
- #7: 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.
- #8: 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.
- #10: 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.
- #11: 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.
- #12: 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 °.
- #13: 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.
- #14: 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
- #16: 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).
- #18: Figure 3-33. Answer: 40.4 ° (simple geometry)
- #19: 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
- #20: Figure 3-35. Answer: 15.7 m/s