Projectile Motion

1. Projectile Motion Chapter 5

2. Objectives • Distinguish between a vector quantity and a scalar quantity. • Explain how to find the resultant of 2 perpendicular vectors. • Describe how the components of a vector affect each other. • Describe the components of projectile motion. • Describe the downward motion of a horizontally launched projectile. • Describe how far below an imaginary straight-line path a projectile falls.

3. 5.1 Vector and Scalar Quantities • Recall from Ch. 2 • A vector quantity is a quantity which is fully described by both magnitude and direction. Velocity and acceleration are vector quantities. • A scalar quantity is a quantity which is fully described by its magnitude. Speed is a scalar quantity. • Scalars can be added, subtracted, divided, and multiplied like any other number.

4. 5.2 Velocity Vectors • Recall (from Ch. 2) that vectors have a head and tail. • When drawing a vector, a scale is developed so that the magnitude of the vector in the diagram is represented by the length of the arrow. • The arrow is drawn a precise length in accordance with a chosen scale.

5. 5.2 Velocity Vectors • The direction of a vector is expressed as the angle of rotation of the vector about its “tail" from either east, west, north, or south. • 30o west of north N W E S

6. 5.2 Velocity Vectors • Velocity of something is often the result of combining 2 or more velocities. • An airplane usually encounters a wind (air which is moving with respect to an observer on the ground below). • A motor boat in a river encounters a river current (water which is moving with respect to an observer on dry land).

7. 5.2 Velocity Vectors • The vectors are scaled to represent 100 km/hr and 25 km/hr. • Since these vectors are in the same direction, they are “in parallel”. • Recall (from Ch. 2) that if vectors are in parallel, the magnitude values ADD.

8. 5.2 Velocity Vectors • The velocity vectors are now in opposite directions. • The magnitude values will, therefore, SUBTRACT.

11. 5.2 Velocity Vectors • Consider now the case of the crosswind (or cross current). • The resultant (the result of adding these 2 vectors) is the DIAGONAL of the rectangle created by the 2 vectors. (Remember the “Pythagorean Theorem” ?)

12. 5.2 Velocity Vectors • Since the two vectors to be added - the southward plane velocity and the westward wind velocity - are at right angles to each other, the Pythagorean theorem can be used to determine the magnitude of the resultant velocity. • 252 + 1002 = r2 • r = √10625 = 103.1 km/hr.

13. Practice • How fast will a boat move upstream if it has a speed of 20 mph and the current downstream is 5 mph? • How fast will the boat move directly across stream if its speed is 16 mph and the current downstream is 12 mph? • Suppose an airplane flying at 80 km/hr encounters a crosswind of 15 km/hr. Will the airplane fly faster or slower than 80 km/hr?

14. 5.3 Components of Vectors • A single vector will often need to be transformed into 2 vectors that are at right angles to each other. • The 2 vectors at right angles that add up to the given vector are known as the components of the vector. The perpendicular components of a vector are independent of each other. • The process of determining the components of a vector is known as resolution. • There are 2 methods of vector resolution - a graphical method and a trigonometric method.

15. Problem: Graphical Vector Resolution • If Fido's dog chain is stretched upward and rightward and pulled tight by his master, then the force in the chain has two components - an upward and a rightward component.

16. Graphical Vector Resolution Vector representing force on Fido • First, using a scale, draw the vector you want to resolve. • Next, draw in x and y axes at the tail of the vector you want to resolve. • For the rightward component, draw a dotted line from the end of the vector, parallel to the y-axis, to its intersect on the x axis. • Indicate the rightward component by drawing a vector from the origin to the dotted line. Rightward component

17. Graphical Vector Resolution • For the upward component, draw a dotted line from the end of the vector, parallel to the x-axis, to its intersect on the y axis. • Indicate the upward component by drawing a vector from the origin to the dotted line. • Measure (using your scale) both components. Upward component

18. Practice • Graphically resolve a velocity vector of 100 km/hr that is at an angle of 600 to the x-axis. • Graphically resolve a force vector of 350 N that is at an angle of 25 0to the x-axis.

19. Problem: Trigonometric Vector Resolution • Construct a sketch (no scale needed) of the vector in the indicated direction; label its magnitude and the angle which it makes with the horizontal. • Draw a rectangle about the vector such that the vector is the diagonal of the rectangle; draw the components of the vector and label them.

20. Trigonometric Vector Resolution 3. a. With the given vector (c in the diagram below) and angle (θ below), use the sine function to solve for the x component (b in diagram). b. Use the cosine function to solve for the y component (a in diagram).

21. Trigonometric Vector Resolution Fvert Fhoriz

22. Practice • A ball is thrown upwards at an angle of 600 from the ground. If its initial velocity is 40 m/s, what is the horizontal component of the velocity? The vertical component? • Carrie digs a hole in her backyard to plant the rose bush her kids gave her for Mother’s Day. If she exerts a force of 30. N on the shovel at an angle of 400to the ground, what are the horizontal and vertical components of Carrie’s force?

23. 5.4 Projectile Motion • A projectile is any object that moves through air or space acted on only by gravity and air resistance.

24. 5.4 Projectile Motion • There are two components of any projectile's motion – the horizontal component and the vertical component.

25. Horizontal Component • The horizontal component is just like the horizontal motion of a ball rolling freely along a level surface with no friction. • With no friction, the horizontal velocity is constant. (v = d/t) • With no outside forces, the horizontal acceleration is 0.

26. Vertical Component • The vertical component is like the motion of a freely falling object. • A projectile’s velocity increases as it falls - it accelerates at about 10 m/s2. (v = gt) • In each time interval, the projectile covers a greater distance. (d = 1/2gt2)

27. 5.4 Projectile Motion The horizontal and vertical components of the motion are INDEPENDENT of each other.

28. Practice Problem • Gerrie rolls a marble off a table. If the marble hits the floor at a distance that is 0.40 m from the table’s edge exactly 0.50 seconds later, what is . . . • Horizontal component of the marble’s velocity? • The vertical velocity at touchdown? • The height of the table?

29. 5.5 Projectiles Launched Horizontally

30. 5.5 Projectiles Launched Horizontally • Notice (in the previous slide) that the horizontal motion remains constant. The canon ball moves the same distance in each second because there are no forces acting on it. • Gravity acts downward so the only acceleration that occurs is downward. The downward motion of a “launched” projectile is the SAME as that of free fall.

31. 5.5 Projectiles Launched Horizontally • The path traced by a projectile is a parabola; the curved path is considered parabolic. • The trajectory (path) is a combination of constant horizontal motion and accelerated vertical motion.

32. Practice Problem • George is working on his house. From a ladder, he tosses his hammer to the ground, 2.8 m below his position. The hammer lands 1.5 m horizontally from George. • How long is the hammer in the air? • What is the horizontal velocity component of George’s throw? • If George’s initial throw has no vertical component, what is the vertical velocity of the hammer when it hits the ground? • What is the velocity of the hammer when it strikes the ground?

33. 5.6 Projectiles Launched at an Angle

34. 5.6 Projectiles Launched at an Angle

35. 5.6 Projectiles Launched at an Angle 5 s 4 s • This diagram shows that there are specific vertical distances for the cannonball from the position of where it should be (with no gravity) to where it actually is (with gravity). • These distances are the SAME distances it would fall if it were dropped and were undergoing free fall for the given amount of time. 3 s 2 s 1 s 5 m 20m 45m 80m 125m

36. 5.6 Projectiles Launched at an Angle • Recall (from Ch. 4) that the distance is given by d = 1/2at2. • Since a = g = 9.8 m/s2, d = 4.9t2. • No matter the angle at which a projectile is launched, the vertical distance of fall from an idealized straight-line path will always be the same for equal times.

38. Practice Problem • Mario, the amazing human cannonball, is fired out of a cannon at a speed of 20. m/s at an angle of 430 to the horizontal. • What are the horizontal and vertical components of his initial speed? • If he lands 52 m away from the canon, how long was he in the air? • Assuming the canon is at ground level, what vertical height did Mario achieve?

39. Height • The vertical distance a projectile falls below an imaginary straight-line path increases continually with time and is equal to 4.9t2 meters.

40. Height • The diagram above shows the path of a projectile with velocity vectors drawn in for different points on its parabolic path. The vectors have been resolved into their x and y components.

41. Height • The horizontal component is always the same – the ball moves equal distances in equal time intervals. • Note that ONLY the vertical component changes – it decreases going upward against gravity and increases going downward with gravity. The projectile accelerates due to Earth’s gravity. • Is the velocity of the projectile at its highest point zero?

42. Range • Now suppose 2 projectiles are launched at the same speed but different angles. The projectile launched at the greater angle will have a velocity vector with a greater vertical component and a smaller horizontal component. • The greater vertical component results in a higher path; the smaller horizontal component results in less range (distance).

44. Horizontal Ranges • Here the paths of 3 cannonballs are shown – all launched at the same initial speed but different projection angles. • Note: • The paths are all parabolas. • Each ball reaches a different height. • What’s going on with the horizontal ranges?

45. Horizontal Ranges • Different projection angles – 300 and 600, for example – can have the same range. This always works out for angles that add up to 900. • For angles less than 450, the smaller the angle, the shorter the horizontal range. • Maximum range is usually attained at an angle of 450.

46. Speed • The cases of projectile motion mentioned in class assume air resistance is negligible. • The speed lost by the projectile going up equals the speed gained while it is coming down. • The time to go up equals the time to come down. • With air resistance, the range of the projectile is diminished and the path is not a true parabola.

47. Speed • Also, the projectile motion discussed in class has been “short-range” projectile motion. For this motion, we assume the ground is flat - the curvature of the Earth’s surface has no effect on the motion. • For “long-range” projectiles, the curvature of the Earth must be taken into account. If a projectile has enough initial velocity, it will “fall all the way around the Earth” and become a satellite.