Presentation is loading. Please wait.

Presentation is loading. Please wait.

Motion in Two Dimensions

Similar presentations

Presentation on theme: "Motion in Two Dimensions"— Presentation transcript:

1 Motion in Two Dimensions
Chapter 4 Motion in Two Dimensions

2 Projectile Motion An object may move in both the x and y directions simultaneously. This form of two-dimensional motion we will deal with is called projectile motion. Section 4.3

3 Assumptions of Projectile Motion
The free-fall acceleration is constant over the range of motion. It is directed downward. This is the same as assuming a flat Earth over the range of the motion. It is reasonable as long as the range is small compared to the radius of the Earth. The effect of air friction is negligible. With these assumptions, an object in projectile motion will follow a parabolic path. This path is called the trajectory. Launched over level ground we see that the symmetry is apparent. Section 4.3

4 Velocity of a projectile
x y v v v vo I must resolve this into components though…. vf Velocity is tangent to the path for the entire trajectory.

5 Projectile Motion Diagram
Notice how the vertical velocity vector changes over the path while the horizontal velocity vector remains constant. Maximum speed is attained at the beginning, and again at the end, of the trajectory if the projectile is launched over level ground. Perpendicular components of motion are independent of one another. Section 4.3

6 Acceleration of a projectile
x y g g g g g Acceleration points down at 9.8 m/s2 for the entire trajectory of all projectiles.

7 Maximum height/Range of a projectile
y Maximum Height The vertical velocity component is zero at maximum height. For a projectile launched over level ground, the maximum height occurs halfway through the flight of the projectile. We will see equations for these in a moment. Range The MAXIMUM HEIGHT of the projectile occurs when it stops moving upward. The RANGE is how far the projectile moves horizontally.

8 Horizontal Component of Velocity
Is constant Not accelerated Not influenced by gravity. (Why?) Will follow the following equation: Δx = vixt Where did we get this from?

9 Vertical Component of Velocity
Undergoes accelerated motion Accelerated specifically by gravity (9.8 m/s2) Vy = Vo,y - gt y = yo + Vo,yt - 1/2gt2 Vy2 = Vo,y2 - 2g(y – yo)

10 Acceleration at the Highest Point
The vertical velocity is zero at the top. What is the acceleration at the top of the trajectory/ maximum height? The acceleration is not zero anywhere along the trajectory. If the projectile experienced zero acceleration at the highest point, its velocity at the point would not change. The projectile would move with a constant horizontal velocity from that point on. Section 4.3

11 Analyzing Projectile Motion
Consider the motion as the superposition of the motions in the x- and y-directions. The actual position at any time is given by: The initial velocity can be expressed in terms of its components. vxi = vi cos q and vyi = vi sin q The x-direction has constant velocity. ax = 0 The y-direction is free fall. ay = -g Section 4.3

12 Projectile Motion Vectors
The final position is the vector sum of the initial position, the position resulting from the initial velocity and the position resulting from the acceleration. Section 4.3

13 Range and Maximum Height of a Projectile
When analyzing projectile motion, two characteristics are of special interest The range, R, is the horizontal distance of the projectile. The maximum height the projectile reaches is h. Section 4.3

14 Height of a Projectile, equation
The maximum height of the projectile can be found in terms of the initial velocity vector: This equation is valid only for symmetric motion. Section 4.3

15 Range of a Projectile, equation
The range of a projectile can be expressed in terms of the initial velocity vector: This is valid only for symmetric trajectory. Section 4.3

16 More About the Range of a Projectile
Section 4.3

17 Range of a Projectile, final
The maximum range occurs at qi = 45o . Complementary angles will produce the same range. The maximum height will be different for the two angles. The times of the flight will be different for the two angles. Section 4.3

18 Deriving the Parabolic Path of a Trajectory

19 Sample Problem The Zambezi River flows over Victoria Falls in Africa. The falls are approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops.

20 Sample Problem An astronaut on the planet Zircon tosses a rock horizontally with a speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a horizontal distance of 8.95 m from the astronaut. What is the acceleration due to gravity on Zircon?

21 Should be able to do this on your own!
Sample Problem Playing shortstop, you throw a ball horizontally to the second baseman with a speed of 22 m/s. The ball is caught by the second baseman 0.45 s later. How far were you from the second baseman? What is the distance of the vertical drop? Should be able to do this on your own!

22 Sample – Non-Horizontally Launched Projectile (Standard Problem)
An object is thrown off a cliff from 50m high with an initial velocity of 25 m/s and an angle of elevation of 30 degrees. How far away from the base of the cliff will it land? (Use 10 m/s2 as g)

23 Sample – Maximum Projectile Range
In order to calculate the maximum range of a rocket, you fire the rocket straight up and record the time it takes for it to return to the ground, ttrajectory. Based on this single piece of data, a. What is the speed with which the rocket is launched? b. Compute the rocket’s range as a function of the angle of elevation, θ and the initial speed, vo. c. At what angle is the range maximized?

24 Sample – Extra Practice
A firefighter, a distance d from a burning building, directs a stream of water from a fire hose at angle θi above the horizontal. If the initial speed of the stream is vi, at what height h does the water strike the building?

25 Projectile Motion – Problem Solving Hints
Conceptualize Establish the mental representation of the projectile moving along its trajectory. Categorize Confirm air resistance is neglected. Select a coordinate system with x in the horizontal and y in the vertical direction. Analyze If the initial velocity is given, resolve it into x and y components. Treat the horizontal and vertical motions independently. Section 4.3

26 Projectile Motion – Problem Solving Hints, cont.
Analysis, cont. Analyze the horizontal motion with the particle-under-constant-velocity model. Analyze the vertical motion with the particle-under-constant-acceleration model. Remember that both directions share the same time. Finalize Check to see if your answers are consistent with the mental and pictorial representations. Check to see if your results are realistic. Section 4.3

27 Non-Symmetric Projectile Motion
Follow the general rules for projectile motion. Break the y-direction into parts. up and down or symmetrical back to initial height and then the rest of the height Apply the problem solving process to determine and solve the necessary equations. May be non-symmetric in other ways Section 4.3

Download ppt "Motion in Two Dimensions"

Similar presentations

Ads by Google