MOTION OF A PROJECTILE Today’s Objectives: Students will be able to: Analyze the free-flight motion of a projectile. In-Class Activities: Check Homework Reading Quiz Applications Kinematic Equations for Projectile Motion Concept Quiz Group Problem Solving Attention Quiz

The downward acceleration of an object in free-flight motion is READING QUIZ The downward acceleration of an object in free-flight motion is A) zero. B) increasing with time. C) 9.81 m/s2. D) 9.81 ft/s2. 2. The horizontal component of velocity remains _________ during a free-flight motion. A) zero B) constant C) at 9.81 m/s2 D) at 32.2 ft/s2 Answers: C B

APPLICATIONS A good kicker instinctively knows at what angle, q, and initial velocity, vA, he must kick the ball to make a field goal. For a given kick “strength”, at what angle should the ball be kicked to get the maximum distance?

APPLICATIONS (continued) A basketball is shot at a certain angle. What parameters should the shooter consider in order for the basketball to pass through the basket? Distance, speed, the basket location, … anything else?

APPLICATIONS (continued) A firefighter needs to know the maximum height on the wall she can project water from the hose. What parameters would you program into a wrist computer to find the angle, q, that she should use to hold the hose?

MOTION OF A PROJECTILE (Section 12.6) Projectile motion can be treated as two rectilinear motions, one in the horizontal direction experiencing zero acceleration and the other in the vertical direction experiencing constant acceleration (i.e., from gravity).

MOTION OF A PROJECTILE (Section 12.6) For illustration, consider the two balls on the left. The red ball falls from rest, whereas the yellow ball is given a horizontal velocity. Each picture in this sequence is taken after the same time interval. Notice both balls are subjected to the same downward acceleration since they remain at the same elevation at any instant. Also, note that the horizontal distance between successive photos of the yellow ball is constant since the velocity in the horizontal direction is constant.

KINEMATIC EQUATIONS: HORIZONTAL MOTION Since ax = 0, the velocity in the horizontal direction remains constant (vx = vox) and the position in the x direction can be determined by: x = xo + (vox) t No air resistance is assumed! Why is ax equal to zero (what assumption must be made if the movement is through the air)?

KINEMATIC EQUATIONS: VERTICAL MOTION Since the positive y-axis is directed upward, ay = – g. Application of the constant acceleration equations yields: vy = voy – g t y = yo + (voy) t – ½ g t2 vy2 = voy2 – 2 g (y – yo) Only any two of the three are independent, based on their derviation. For any given problem, only two of these three equations can be used. Why?

Find: Horizontal distance it travels and vC. Plan: EXAMPLE I Given: vA and θ Find: Horizontal distance it travels and vC. Plan: Apply the kinematic relations in x- and y-directions. Solution: Using vAx = 10 cos 30 and vAy = 10 sin 30 We can write vx = 10 cos 30 vy = 10 sin 30 – (9.81) t x = (10 cos 30) t y = (10 sin 30) t – ½ (9.81) t2 Source : F12-22 Since y = 0 at C 0 = (10 sin 30) t – ½ (9.81) t2  t = 0, 1.019 s

EXAMPLE I (continued) Only the time of 1.019 s makes sense! Velocity components at C are; vCx = 10 cos 30 = 8.66 m/s  vCy = 10 sin 30 – (9.81) (1.019) = -5 m/s = 5 m/s  v C = 8.66 2 + (−5) 2 =10 m/s Horizontal distance the ball travels is; x = (10 cos 30) t x = (10 cos 30) 1.019 = 8.83 m

Given: Projectile is fired with vA=150 m/s at point A. EXAMPLE II Given: Projectile is fired with vA=150 m/s at point A. Find: The horizontal distance it travels (R) and the time in the air. Plan: How will you proceed? Source : F12-26

EXAMPLE II Given: Projectile is fired with vA=150 m/s at point A. Find: The horizontal distance it travels (R) and the time in the air. Plan: Establish a fixed x, y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x- and y-directions.

EXAMPLE II (continued) Solution: 1) Place the coordinate system at point A. Then, write the equation for horizontal motion. +  xB = xA + vAx tAB where xB = R, xA = 0, vAx = 150 (4/5) m/s Range, R, will be R = 120 tAB 2) Now write a vertical motion equation. Use the distance equation. + yB = yA + vAy tAB – 0.5 g tAB2 where yB = – 150, yA = 0, and vAy = 150(3/5) m/s We get the following equation: –150 = 90 tAB + 0.5 (– 9.81) tAB2 Solving for tAB first, tAB = 19.89 s. Then, R = 120 tAB = 120 (19.89) = 2387 m

A) (vo sin q)/g B) (2vo sin q)/g C) (vo cos q)/g D) (2vo cos q)/g CONCEPT QUIZ 1. In a projectile motion problem, what is the maximum number of unknowns that can be solved? A) 1 B) 2 C) 3 D) 4 2. The time of flight of a projectile, fired over level ground, with initial velocity Vo at angle θ, is equal to? A) (vo sin q)/g B) (2vo sin q)/g C) (vo cos q)/g D) (2vo cos q)/g Answers: C B

GROUP PROBLEM SOLVING I x y Given: A skier leaves the ski jump ramp at qA = 25o and hits the slope at B. Find: The skier’s initial speed vA. Plan: Source : P12-96

GROUP PROBLEM SOLVING I Plan: Establish a fixed x,y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x- and y-directions. x y Given: A skier leaves the ski jump ramp at qA = 25o and hits the slope at B. Find: The skier’s initial speed vA.

GROUP PROBLEM SOLVING I (continued) Solution: Motion in x-direction: Using xB = xA + vox(tAB)  (4/5)100 = 0 + vA (cos 25) tAB = tAB= 80 vA (cos 25) 88.27 vA – 64 = 0 + vA(sin 25) { } 88.27 vA – ½ (9.81) { }2 Motion in y-direction: Using yB = yA + voy(tAB) – ½ g(tAB)2 vA = 19.42 m/s tAB= (88.27 / 19.42) = 4.54 s

GROUP PROBLEM SOLVING II Given: The golf ball is struck with a velocity of 80 ft/s as shown. Find: Distance d to where it will land. Plan: x y Source : P12-93

GROUP PROBLEM SOLVING II Given: The golf ball is struck with a velocity of 80 ft/s as shown. Find: Distance d to where it will land. x y Plan: Establish a fixed x, y coordinate system (in this solution, the origin of the coordinate system is placed at A). Apply the kinematic relations in x- and y-directions.

GROUP PROBLEM SOLVING II (continued) Solution: x y Motion in x-direction: Using xB = xA + vox(tAB) d cos10 = 0 + 80 (cos 55) tAB tAB = 0.02146 d Motion in y-direction: Using yB = yA + voy(tAB) – ½ g(tAB)2 d sin10 = 0 + 80(sin 55)(0.02146 d) – ½ 32.2 (0.02146 d)2 0 = 1.233 d – 0.007415 d2 d = 0, 166 ft Only the non-zero answer is meaningful.

ATTENTION QUIZ 1. A projectile is given an initial velocity vo at an angle f above the horizontal. The velocity of the projectile when it hits the slope is ____________ the initial velocity vo. A) less than B) equal to C) greater than D) None of the above. 2. A particle has an initial velocity vo at angle f with respect to the horizontal. The maximum height it can reach is when A) f = 30° B) f = 45° C) f = 60° D) f = 90° Answers: 1. A 2. D

End of the Lecture Let Learning Continue