1. Copyright © 2010 Pearson Education, Inc. Lecture Outline Chapter 4 Physics, 4 th Edition James S. Walker

2. Copyright © 2010 Pearson Education, Inc. Chapter 4 Two-Dimensional Kinematics

3. Copyright © 2010 Pearson Education, Inc. Units of Chapter 4 Motion in Two Dimensions Projectile Motion: Basic Equations Zero Launch Angle General Launch Angle Projectile Motion: Key Characteristics

4. Copyright © 2010 Pearson Education, Inc. 4-1 Motion in Two Dimensions If velocity is constant, motion is along a straight line:

5. Copyright © 2010 Pearson Education, Inc. 4-1 Motion in Two Dimensions An eagle perched on a tree limb 19.5 m above the water spots a fish swimming near the surface. The eagle pushes off from the branch and descends toward the water. By adjusting its body in flight, the eagle maintains a constant speed of 3.10 m/s at an angle of 20.0 0 below the horizontal. (a) How long does it take for the eagle to reach the water? (b) How far has the eagle traveled in the horizontal direction when it reaches the water? V ox V oy V ox = v 0 cos θ = (3.10 m/s)cos 20.0 0 = 2.91 m/s V oy = -v 0 sin θ = -(3.10 m/s)sin 20.0 0 = -1.06 m/s Y = y o + v oy (t) = 0 (a). t = - h/v oy = - 19.5 m / (-1.06 m/s) = 18.4 s (b). X = X 0 + V ox (t) = 0 + (2.91 m/s)(18.4 s) = 53.5 m

6. Copyright © 2010 Pearson Education, Inc. 4-1 Motion in Two Dimensions Motion in the x- and y-directions should be solved separately:

7. Copyright © 2010 Pearson Education, Inc. 4-2 Projectile Motion: Basic Equations A hummingbird is flying in such a way that it is initially moving vertically with a speed of 4.6 m/s and accelerating horizontally at 11 m/s 2. Assuming the bird’s acceleration remains constant for the time interval of interest, find (a) the horizontal and vertical distances through which it moves in 0.55 s and (b) its x and y velocity components at t = 0.55s. (a). X = x 0 + v ox t + 1/2 a x t 2 = 0 + 0 + ½(11 m/s 2 ) (0.55 s) 2 = 1.7 m y = y 0 + v oy t + 1/2 a y t 2 = 0 + ½(4.6 m/s 2 ) (0.55 s) 2 + 0 = 2.5 m (b) V x = v 0x + a x t = 0 + (11 m/s 2 )(0.55 s) = 6.1 m/s V y = v 0y + a y t = 4.6 m/s + (0 m/s 2 )(0.55 s) = 4.6 m/s The bird’s velocity at 0.55 s is V = (v x 2 + v y 2 ) 1/2 = ((6.1 m/s) 2 + (4.6 m/s) 2 ) 1/2 = 7.6 m/s At an angle of θ => θ = tan-1 (v y /v x ) = tan -1 [(4.6 m/s) / (6.1 m/s)] = 37 0

8. Copyright © 2010 Pearson Education, Inc. 4-2 Projectile Motion: Basic Equations In studying projectile motion we made the following assumptions: ignore air resistance the acceleration due to gravity is constant, downward, and has a magnitude equal to g = 9.81 m/s 2, downward the Earth’s rotation is ignored If y-axis points upward, acceleration in x-direction is zero and acceleration in y-direction is -9.81 m/s 2. Thus, the x component of acceleration is zero. a x = 0

9. Copyright © 2010 Pearson Education, Inc. 4-2 Projectile Motion: Basic Equations Suppose, the x axis is horizontal and the y axis is vertical, with the positive direction upward. Since downward is the negative direction, it follows that a y = -9.81 m/s 2 = -g Gravity causes no acceleration in the x direction. Thus, the x component of acceleration is zero : a x = 0 With these acceleration components substituted into the fundamental constant- acceleration equations of motion (Table 4-1) we find: These, then, are the basic equations of projectile motion: Projectile Motion (a x = 0, a y = -g)

10. Copyright © 2010 Pearson Education, Inc. 4-3 Zero Launch Angle Launch angle: direction of initial velocity with respect to horizontal Launch angle of a projectile (a)A projectile launched at an angle above the horizontal, θ. 0. A launch below the horizontal would correspond to θ < 0. (b) A projectile launched horizontally, θ = 0. In this section we consider θ = 0.

11. Copyright © 2010 Pearson Education, Inc. 4-3 Zero Launch Angle In this case, the initial velocity in the y-direction is zero. Here are the equations of motion, with x 0 = 0 and y 0 = h: Substituting these specific into the fundamental equations for projectile motion (Equation 4-6 gives the following simplified results for zero launch angle (θ = 0) :

12. Copyright © 2010 Pearson Education, Inc. 4-3 Zero Launch Angle This is the trajectory of a projectile launched horizontally: In this plot, the projectile was launched from a height of 9.5 m with an initial speed of 5.0 m/s. The positions shown in the plot correspond to the times t = 0.2 s, 0.4 s, 0.60 s………

13. Copyright © 2010 Pearson Education, Inc. 4-3 Zero Launch Angle A person skateboarding with a constant speed of 1.30 m/s releases a ball from a height of 1.25 m above the ground. Given that x o = 0 and y 0 = h 1.25 m, find x and y for (a) t = 0.25 s and (b) t = 0.5 s. ( c) Find the velocity, speed and direction of motion of the ball at t = 0.50 s. The ball starts at x 0 = 0 and y 0 = h = 1.25 m (a)Substitute t = 0.25 s into the x and y equation of motion. x = v 0 t = (1.30 m/s)(0.25 s) = 0.325 m y = h – 1/2gt 2 = 1.25 m – ½(9.81 m/s 2 )(0.25 s) 2 = 0.943 m (b) Substitute t = 0.5 s into the x and y equations of motion. x = v 0 t = (1.30 m/s)(0.50 s) = 0.650 m y = h – 1/2gt 2 = 1.25 m – ½(9.81 m/s 2 )(0.50 s) 2 = 0.0238 m (c ) Calculate the x and y components of the velocity at t = 0.50 s using V x = V 0 and V y = -gt V x = V 0 = 1.30 m/s V y = -gt = -(9.81 m/s 2 )(0.50 s) = -4.91 m/s V = (1.30 m/s)x + (-4.91 m/s)y = > V = (V x 2 + V y 2 ) 1/2 V = ((1.3 m/s) 2 + (-4.91 m/s) 2 ) 1/2 = 5.08 m/s At an angle of θ => θ = tan-1 (v y /v x ) = tan -1 [(-4.91 m/s) / (1.30 m/s)] = -75.2 0

14. Copyright © 2010 Pearson Education, Inc. 4-3 Zero Launch Angle What is the shape of the curved path followed by a projectile launched horizontally? This can be found by combining x = v o t and y = h –gt 2. t = x/v 0 Eliminating t and solving for y as a function of x: This has the form y = a + bx 2, which is the equation of a parabola where a = h = constant and b = -g/2v 0 2 = constant.. Where does a projectile land if it is launched horizontally with a speed v0 from a height h? The landing point can be found by setting y = 0 and solving for x: 0 = h – (g/2v 0 2 )x 2

15. Copyright © 2010 Pearson Education, Inc. 4-4 General Launch Angle In general, v 0x = v 0 cos θ and v 0y = v 0 sin θ This gives the equations of motion:

16. Copyright © 2010 Pearson Education, Inc. 4-4 General Launch Angle Snapshots of a trajectory; red dots are at t = 1 s, t = 2 s, and t = 3 s

17. Copyright © 2010 Pearson Education, Inc. 4-5 Projectile Motion: Key Characteristics Range: the horizontal distance a projectile travels If the initial and final elevation are the same:

18. Copyright © 2010 Pearson Education, Inc. 4-5 Projectile Motion: Key Characteristics The range is a maximum when θ = 45°:

19. Copyright © 2010 Pearson Education, Inc. Symmetry in projectile motion: 4-5 Projectile Motion: Key Characteristics

20. Copyright © 2010 Pearson Education, Inc. Summary of Chapter 4 Components of motion in the x- and y- directions can be treated independently In projectile motion, the acceleration is –g If the launch angle is zero, the initial velocity has only an x-component The path followed by a projectile is a parabola The range is the horizontal distance the projectile travels