 # CHAPTER 3 PROJECTILE MOTION. North South EastWest positive x positive y negative x negative y VECTORS.

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CHAPTER 3 PROJECTILE MOTION

North South EastWest positive x positive y negative x negative y VECTORS

VECTOR EXAMPLE 1 X-component X-component10 Y-component Y-component0 10 m East 10 m

VECTOR EXAMPLE 2 X-component X-component-10 Y-component Y-component0 10 m West 10 m

VECTOR EXAMPLE 3 X-component X-component0 Y-component Y-component10 10 m North 10 m

VECTOR EXAMPLE 4 X-component X-component0 Y-component Y-component-10 10 m South 10 m

COMPONENTS OF A VECTOR V VXVX VYVY x y 

EXAMPLE A wind with a velocity of 40 m/s blows towards 30  NE. What are the x and y components of the wind’s velocity. VxVx VyVy V=40 

PROJECTILE A projectile is an object with an initial velocity that is allowed to move under the affects of gravity. Ex. a javelin throw, a package released by an airplane, a thrown baseball

Types of Projectiles

Trajectory: is the path of the projectile. Because of Earth’s Gravitational pull and their own inertia, projectiles follow a curved path. They have both horizontal and vertical velocities.

Constant horizontal velocity due to inertia V ix = V fx a x = 0

Constant horizontal velocity due to inertia V iY = 0 a Y = -9.8 m/s 2

IMPORTANT!!! The ball’s horizontal and vertical motions are completely independent of each other.

CURVED MOTION UNDER GRAVITY 1. Ball is simply dropped 2. Ball is thrown horizontally at 10 m/s 3. Ball is thrown horizontally at 20 m/s 1 2 3

In the absence of any frictional forces (like air), all the three balls fall to the ground at the same time. The horizontal motion does not affect the vertical acceleration. All the three balls are pulled to the ground in the same way because of gravity. All the three balls travel the same vertical distance in the same time.

PROJECTILES LAUNCHED HORIZONTALLY INITIAL VELOCITY RANGE (D X ) TRAJECTORY HEIGHT (D Y )

X-component initial velocity horizontal distance (range) zero acceleration final velocity= initial velocity Y- component zero initial velocity negative vertical distance -9.8 m/s 2 acceleration Negative final velocity

PROBLEM 1 A stone is thrown horizontally at 15 m/s from the top of a cliff 44 m high. A) How long does the stone take to reach the ground? B) How far from the base of the cliff does the stone strike the ground? C) Sketch the trajectory of the stone.

X-component V i = 15 m/s a = 0 m/s 2 V f =15 m/s Y- component V i = 0 m/s a = -9.8 m/s 2 D = - 44 m

X-component V i = 15 m/s a = 0 m/s 2 V f =15 m/s Y- component V i = 0 m/s a = -9.8 m/s 2 D = - 44 m D = V i t + ½ at 2 - 44 = ½ (-9.8)t 2 t = 3 s D = V i t + ½ at 2 D = 15(3) = 45m

A stone is thrown horizontally at a speed of + 5 m/s from the top of a cliff 78.4 m high. A) How long does the stone take to reach the bottom of the cliff? B) How far from the base of the cliff does the stone strike the ground? C) What are the horizontal and vertical velocities of the stone just before it hits the ground? PROBLEM 2

X-component V i = 5 m/s a = 0 m/s 2 V f =5 m/s Y- component V i = 0 m/s a = -9.8 m/s 2 D = - 78.4 m

X-component V i = 5 m/s a = 0 m/s 2 V f =5 m/s Y- component V i = 0 m/s a = -9.8 m/s 2 D = - 78.4 m D = V i t + ½ at 2 -78.4=½ (-9.8) t 2 t = 4 s D = V i t + ½ at 2 D = 5(4) = 20 m

Y- component V i = 0 m/s a = -9.8 m/s 2 D = - 78.4 m t = 4 s V f = V i + at = 0+(-9.8) 4 = 0+(-9.8) 4 =- 39.2 m/s

How would a) and b) change if the stone is thrown with twice the velocity. Answer: A) would not change B) would increase two times PROBLEM 3

How would a) and b) change if the stone is thrown with the same speed but twice the height. Answer: A) would increase B) would increase PROBLEM 4

A steel ball rolls with a constant velocity across a table top 0.950 m high. It rolls and hits off the ground +.350 m horizontally from the edge of the table. How fast was the ball rolling? PROBLEM 5

X-component V i = ? m/s a = 0 m/s 2 V f =? m/s D = +.350 m Y- component V i = 0 m/s a = -9.8 m/s 2 D = -.950 m

X-component V i = ? m/s = V f a = 0 m/s 2 D =.350 m t =.44 s D = V i t + ½ at 2 0.350 = V i (.44) V i =.80 m/s Y- component V i = 0 m/s a = -9.8 m/s 2 D = -0.950 m D = V i t + ½ at 2 -0.950= ½(-9.8)t 2 t =.44 s

Suppose you throw a ball upward at an angle. The ball has two types of motion a) Horizontal Motion b) Vertical Motion

The ball moves horizontally at a constant speed. As it moves up gravity slows the ball down (the y speed decreases). At the maximum height, the ball stops. It changes direction and falls downward. Gravity increases the speed of the ball as it falls.

The path of motion is an arc- shaped curve known as a parabola. Horizontal motion has no effect on the time it takes an object to fall to the ground.

PROJECTILES LAUNCHED AT AN ANGLE  LAUNCHING VELOCITY LAUNCHING VELOCITY X-COMP LAUNCHING VELOCITY Y-COMP

X-component V iX = V cos  horizontal distance (range) zero acceleration final velocity= initial velocity Y- component V iY = V sin  @ maximum height (y velocity = 0) -9.8 m/s 2 acceleration At the end. D=0 Final Velocity=Negative LAUNCHING VELOCITY = v

The initial velocity of a ball in projectile motion is 4.47 m/s. It is projected at an angle of 66  above the horizontal. Find A) how long did it take to land.? B) how high did the ball fly? C) what was its range? PROBLEM 6

X-component V iX =4.47 cos 66 0 = 1.8 m/s a X = 0 m/s 2 V fX =1.8 m/s Y- component V iY =4.47 sin 66 0 =4.1 m/s a Y =-9.8 m/s 2 66 0 4.47 m/s B A C

X-component V iX =4.47 cos 66  = 1.8 m/s a X = 0 m/s 2 V fX =1.8 m/s Y- component V iY =4.47 sin 66  = 4.1 m/s a Y = -9.8 m/s 2 At B V fY = 0 m/s 2a Y D Y = V fY 2 - V iY 2 D Y =.86 m At B (Y-comp) V fY = V iY + a Y t 0=4.1+(-9.8)t t =.417 s

X-component V iX = 1.8 m/s a X = 0 m/s 2 t =.83 s D X = V iX t+ ½a X t 2 D = (1.8)(.83) D = 1.5 m ANSWERS Time =.83 s Range = 1.5 m Height =.86 m

A long jumper leaves the ground at an angle of 20  to the horizontal and a speed of 11 m/s. How far does he jump? What is the maximum height reached? PROBLEM 7

X-component V iX = 11 cos 20 0 = 10.33 m/s a X = 0 m/s 2 V fX = 10.33 m/s Y- component V iY = 11 sin 20 0 = 3.76 m/s a Y = -9.8 m/s 2 20 0 11 m/s B A C

At B V fY = 0 m/s 2a Y D Y = V fY 2 - V iY 2 D Y =.72 m At B (Y-comp) V fY = V iY + a Y t 0= 3.76+(-9.8)t t =.38 s X-component V iX = 11 cos 20  = 10.33 m/s a X = 0 m/s 2 V fX = 10.33 m/s Y- component V iY = 11 sin 20  = 3.76 m/s a Y = -9.8 m/s 2

X-component V iX = 10.33 m/s a X = 0 m/s 2 t = 2*.38 =.76 s D X = V iX t + ½a X t 2 D X =(10.33) (.76) D X = 7.9 m ANSWERS Time =.76 s Range = 7.9 m Height =.72 m

A player kicks a football from the ground level with a velocity of 27 m/s at an angle of 30  above the horizontal. Find A) the hang time (the time the ball is in the air) B) the distance the ball travels before it hits the ground. C) its maximum height. PROBLEM 8

X-component V i = 27 cos 30  = 23.4 m/s a = 0 m/s 2 V f = 23.4 m/s Y- component V i = 27 sin 30  = 13.5 m/s a = -9.8 m/s 2 30 0 27 m/s B A C

At B V f = 0 m/s 2aD = V f 2 - V i 2 D = 9.3 m V f = V i + at 0=13.5+(-9.8)(t) t = 1.38 s X-component V i = 27 cos 30 0 = 23.4 m/s a = 0 m/s 2 V f = 23.4 m/s Y- component V i = 27 sin 30 0 = 13.5 m/s a = -9.8 m/s 2

X-component V i = 23.4 m/s a = 0 m/s 2 t = 2*1.38 =2.76 s D = V i t + ½at 2 D=(23.4) (2.76) D = 65 m ANSWERS Time = 2.76 s Range = 65.8 m Height = 9.3 m

A kicker now kicks the ball with the same speed but at an angle of 60 0 above the horizontal. Find A) the hang time (the time the ball is in the air) B) the distance the ball travels before it hits the ground. C) its maximum height. PROBLEM 9

X-component V i = 27 cos 60 0 = 13.5 m/s a = 0 m/s 2 V f = 13.5 m/s Y- component V i = 27 sin 60 0 = 23.4 m/s a = -9.8 m/s 2 60 0 27 m/s B A C

At B V f = 0 m/s 2aD = V f 2 - V i 2 D = 28 m V f = V i + at 0=23.4+(-9.8)(t) t = 2.39 s X-component V i = 27 cos 60 0 = 13.5 m/s a = 0 m/s 2 V f = 13.5 m/s Y- component V i = 27 sin 60 0 = 23.4 m/s a = -9.8 m/s 2

X-component V i = 13.5 m/s a = 0 m/s 2 t = 2(2.39) =4.78 s D = V i t + ½at 2 D = (13.5) (4.78) D = 65 m ANSWERS Time = 5.76 s Range = 65 m Height = 28 m

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