Motion in two dimensions

Lesson presentation

1. TWO DIMENSIONAL Motion MPHIRISENI NORMAN KHWANDA 5 March 2013

2. Pre-Concept Questions 1. A projectile is fired into the air, and it follows the parabolic path as shown in the drawing There is no air resistance. At any instant, the projectile has a velocity and an acceleration . Which one or more the drawings on the right could not represent the directions for and at any point on the trajectory? A : Diagram 1 and 2 B : Diagram 3 and 4 C : Diagram 1 and 3 D : Diagram 2 and 4 The gravitational acceleration cannot be towards the horizontal direction and again cannot face up.  

3. The Convertible Car 2. Suppose you are driving in a convertible with the top down. The car is moving to the right with constant velocity as the diagram illustrates. You point a rifle upwards and fire it. In the absence of air resistance, where would the bullet land? A : Behind you (or the car) B : Ahead of the you (car) C : In the barrel of the riffle (back to where it was originally fired) The bullet moves with the same velocity horizontally as the car it only accelerates up and down due to gravity . The vertical and horizontal motions are independent of each other

4. The CASE of Two Balls 3. Ball 1 is thrown into the air and it follows the trajectory for a projectile motion shown in the drawing. At the instant Ball 1 is at the top, Ball 2 is dropped from rest from the same height, which ball reaches the ground first? A: Ball 1 reaches the ground first since it is moving while Ball 2 is stationery B: Ball 2 reaches the ground first because it has a shorter distance to travel C: Both balls reach the ground at the same time D: There is not enough information to tell which ball reaches the ground first Explanation: Both have zero velocity vertically and same height and under the gravitational acceleration g. hence they will reach at the same time

5. The path taken by the ball 4. The diagram represents a ball moving at a constant velocity of 50 m/s towards the right. Complete the path that the ball would take after reaching the edge B . Explanation: The ball will continue to move with the same velocity of 50 m/s horizontally but at the same time the force of gravity will be pulling the ball down immediately after leaving the edge, hence the parabolic path

6. Motion in Two Dimensions Summary : The motion is under the influence of gravity which always points towards the centre of the earth . The vertical and horizontal motions are independent from one another The motion horizontally remains constant (acceleration is zero) Air resistance is ignored The maximum velocity during free-fall depends on the initial velocity and initial position (height) Same height, same initial velocity implies same final velocity

7. Equations of Kinematics Horizontally (along the x)

8. 3.2 Equations of Kinematics vertically (along the y)

9. The combination: Simultaneous motion along the and   The x part of the motion occurs exactly as it would if the y part did not occur at all, and vice versa.

10. Assumptions The motion along the and the are independent of each other. The acceleration horizontally is zero The acceleration vertically is constant and due to gravity The effects of air resistance is ignored.  

11. Equations of Kinematics for Constant Acceleration in Two Dimensional Motion

12. The Final velocity of an object Note : The magnitude of velocity along the doesn’t change and is given by because the acceleration The magnitude of the final velocity is given by and the direction is given by The magnitude of velocity along the y changes because of the acceleration  

13. Note : The magnitude of velocity along the doesn’t change and is given by because the acceleration Hence:   Note : The velocity along the y changes due to the gravitational acceleration g because the acceleration Hence the final velocity along y is  

14. The maximum height H and the range R Maximum Height H: Data: set Then and From the diagram , at maximum height Hence the maximum height can be found using the equation: Substituting: The maximum height H The time to reach the maximum height is given by:   Range R: Data set: and The range R can be given by : since Substituting R , hence R t But due to symmetry Hence the total time is The range R = Or because  

15. Your Turn A football was shot at an angle of above the horizontal with the speed of . Calculate The maximum height. The time the ball travel before it hits the ground How far did the ball travel just before hitting the ground?[ The Range ] The magnitude and direction of velocity at maximum height The magnitude and direction of acceleration at maximum height.  

16. Solution What are the information we know and can deduce or calculate from the statement of the problem .? Direction : Assume up to be positive and For the sake of referencing, set Initial velocities: and . At maximum height, and For maximum height: hence substituting we get Hence maximum height The time is given by: Solving we get corresponds to the initial position and is the total time of travel of the ball. (c) The total distance travelled horizontally is the range R given by So that