Physics

Two-Dimensional

Kinematics

Teacher: Luiz Izola

Chapter Preview

1. Motion in Two Dimensions

2. Projectile Motion Basic Equations

3. Zero Launch Angle

4. General Launch Angle

5. Projectile Motion Key Characteristics

Learning Objectives

Apply the motion formulas in two directions (x, y).

Learn how to calculate motion on projectiles

Learn about circular motion.

Introduction

This presentation will concentrate on the study of motion in 2 dimensions.

Particularly, we will concentrate on the study of projectile motion and circular motion.

Projectile motion involves the study of objects that are initially launched and continue moving under gravity’s influence.

Force and Mass

Chapter’s Main Idea:

“Horizontal and Vertical Motions

are independent.” For example: A ball thrown horizontally with a speed v continues to move with the same speed v in the horizontal direction, even as it falls with an increasing speed in the vertical direction. The time of fall is the same whether a ball is dropped from rest straight down or thrown horizontally. Each motion continues independently of the other.

Motion in Two Dimensions

Motion with constant velocity, determining x an y as functions of time.

Ex: A turtle starts at the origin at t=0 and moves with a constant speed v0 = 0.26m/s as show below. How far the turtle moved in the x, y directions after 5.0secs?

Motion in Two Dimensions

Ex2: An eagle perched on a tree limb 19.5m above the water spots a fish swimming near the surface. The eagle pushes off the branch and descends toward the water. By adjusting its body in flight, the eagle maintains a constant speed of 3.10m/s at an angle of 200 below the horizontal. (a) How long does it take for the eagle to reach the water? (b) How long has the eagle traveled in the horizontal when it reaches the water?

Constant Acceleration

To study motion with constant acceleration in two dimensions, we repeat what was done with one dimension but we create two sets of equations, one set for each dimension (x, y).

Table 4-1Constant-Acceleration Equations of Motion

Position as a function of time

Velocity as a function of time

Velocity as a function of position

x = x0 + v0xt + ½ axt2 vx = v0x + axt vx2 = v0x

2 + 2axx

y = y0 + v0yt + ½ ayt2 vy = v0y + ayt vy2 = v0y

2 + 2ayx

Constant AccelerationEx: A hummingbird is flying in such a way that it is initially

moving vertically with a speed of 4.6m/s and accelerating horizontally at 11m/’s2. Assuming the bird’s acceleration remains constant for the time interval of interest, find the horizontal and vertical distance through which it moves in 0.55s.

Projectile Motion Basic Equations A projectile is an object that is thrown, kicked, batted, or

launched into motion and allowed to follow a path solely determined by the influence of gravity.

In studying projectile motion, we make the following assumptions:

I. Air resistance is ignored

II. The acceleration of gravity, g =9.8m/s2, and is constant and downward.

III. The Earth’s rotation is ignored.

Projectile Motion Basic Equations

Suppose, in the figure below, that the x-axis is the horizontal and the y-axis is the vertical, with the positive direction upward. All objects in free fall have ax = 0 and ay = -g. Since the downward direction is negative, it follows that:

ay = 9.81m/s2 = -g

Projectile Motion Basic Equations Gravity cause no acceleration in the x direction.

ax = 0

With these acceleration components placed into the fundamental acceleration equations of motion, we have:

x = x0 + v0xt vx = v0x vx2 = v0x

2

y = y0 + v0yt - ½gt2 vy = v0y -gt vy2 = v0y

2 - 2gy

Zero Launch Angle A special case is a projectile launched horizontally, so

that the angle between the initial velocity and the horizontal is Θ = 0.

Equations of Motion: Choosing ground level to be y = 0 and the release point

to be directly above the origin, the initial position of the ball is given by:

x0 = 0

y0 = 0

Zero Launch Angle The initial velocity is horizontal, corresponding to Θ = 0,

see figure next slide. As a result, the x component of the initial velocity is simply the initial speed,

v0x = v0cos(Θ) = v0

And the y component:

v0y = v0sin(Θ) = 0

Replacing these values in the motion formulas we have:

x = v0t vx = v0 vx2 = v0

2

y = h - ½gt2 vy = -gt vy2 = - 2gy

Zero Launch Angle

Zero Launch AngleEx: A person walking with a speed of 1.30m/s releases a ball

from a height of 1.25m above the ground. Given that x0 = 0 and y0 = h = 1.25m, find x and y for (a) t = 0.250s and (b) t = 0.50s. (c) Find the velocity, speed, and direction of the ball at t = 0.50s. (d) How long does it take for the ball to land?

Parabolic Path and Landing Site The curved path of the projectile can be found by

combining the following:

x = v0t and h = -gt2

This will give:

y = h – 1/2g(x/v0)x2

Where does a projectile land if it is launched horizontally with speed v0 from a height h?

By setting y = 0 in the above equation, we have:

x = v0 (2h/g)

Parabolic Path and Landing SiteEx: A mountain climber encounters a crevasse in an ice field.

The opposite side of the crevasse is 2.75m lower, and is separated horizontally by a distance of 4.10m. To cross the crevasse, the climber gets a running start and jumps in the horizontal direction. (a) What is the minimum speed needed by the climber to safely cross the crevasse? If, instead, the climber’speed is 6.00m/s, (b) where does the climber land?, and (c) What is the climber’s speed at landing?

Practice Problems

1. A sailboat runs before the wind with a constant speed of 2.8m/s in a direction of 520 north of west. How far (a) west and (b) north has the sailboat traveled in 35min?

2. Starting from rest a car accelerates at 2.0m/s2 up a hill that is inclined 5.50 above the horizontal. How far (a) horizontally, and (b) vertically has the car traveled in 12s?

General Launch Angle Consider now, the case of a projectile launched at an

arbitrary angle with respect to the horizontal. To simplify the resulting equation, we take the launching site to be the origin.

General Launch AngleSince the projectile starts at the origin, we have:

x0 = y0 = 0

The components of the initial velocity are:

v0x = v0cos(Θ)

v0y = v0sin(Θ)

The resulting equations are:

x = v0cos(Θ)t vx = v0cos(Θ) vx2 = v0

2cos2(Θ)

y = v0sin(Θ) t - ½ gt2 vy = v0sin(Θ) - gt vy2 = v0

2sin2(Θ) – 2gy

Practice Problems

1. Chipping from the rough a, a golfer sends the ball over a 3.00m-high tree that is 14.0m away. The ball lands at the same level from which it was struck after traveling an horizontal distance of 17.8m. (a) If the ball left the club 540 above the horizontal and landed on the green 2.24s later, what was its initial speed? (b) How high was the ball when it passed over the tree?

Practice Problems2. A projectile is launched from the origin with an initial speed

of 20.0m/s at an angel of 350 above the horizontal. Find the x,y positions of the projectile at times (a) t=0.50s, (b) t=1.00s, and (c) t = 1.50s.

3. A golfer hits a ball with an initial speed of 30m/s at an angle of 500 above the horizontal. The ball lands on a green that is 5.00m above the level where the ball was struck.

a) How long is the ball in the air?

b) How far has the ball traveled in the horizontal when it lands?

c) What is the speed and direction of motion of the ball just before it lands?

Practice ProblemsA trained dolphin leaps from the water with an initial speed of

12m/s (see figure below). In the absence of gravity, the dolphin would move on a straight line to the ball and catch it, but because of gravity, the dolphin follows a parabolic path well below the ball’s initial position. If the trainer releases the ball the instant the dolphin leaves the water, show where the dolphin and falling ball meet.

Projectile Motion: Key CharacteristicsRANGE

The range, R, of a projectile is the horizontal distance it travels before landing. Considering initial and final elevations the same (y = 0).

Challenge: Using the assumption about y = 0 above and the following formulas of motion:

y = v0sin(Θ) t - ½ gt2

x = v0cos(Θ)t

Prove that R is equals to:

R = (v02/g ).sin2Θ

Practice Problems1) A football game begins with a kickoff in which the ball

travels a horizontal distance of 41m. If the ball is kicked at an angle 400 above the horizontal, what was its initial speed?

2) A golf ball is struck on level ground. It lands 92.2m away 4.30s later. What was the direction and magnitude of the initial velocity?

3) The men’s world’s record for the shot put, 23.12m, was set by Randy Barnes of the USA on May 20, 1990. If the shot was launched from 1.80m above the ground at an initial angle of 420, what was the initial speed?

Maximum HeightChallenge: Using the two formulas below

y = v0sin(Θ) t - ½ gt2

vy = v0y - gt

Prove that the maximum height can be calculate as:

h = v0sin(Θ) / 2g

Practice ProblemsThe archerfish hunts by dislodging an unsuspecting insect from

its resting place with a stream of water expelled from the fish’s mouth. Suppose the archerfish squirts water with an initial speed of 2.30m/s at an angle of 19.50 above the horizontal. When the stream of water reaches a beetle on the leaf at height h above the water’s surface it is moving horizontally. (a) How much time does the beetle have to react? (b) What is the beetle’s height? (c) What the horizontal distance between the beetle and the fish?

HomeworkI. Page 102: 11, 13, 15, 17

II. Page 103: 19, 21, 23

III. Page 104: 25

Note: It would be excellent to try all the projectile motion problems in order to get more proficient.