1. Chapter 4, Motion in 2 Dimensions

2. Position, Velocity, Acceleration
Just as in 1d, in 2d, an object’s motion is completely known if it’s position, velocity, & acceleration are known.
Position Vector  r
In terms of unit vectors discussed last time, for an object at position (x,y) in x-y plane:
r  x i + y j
For an object moving: r depends on time t:
r = r(t) = x(t) i + y(t) j

3. Δt = tf - ti vavg  (Δr/Δt) The Displacement Vector is:  Δr = rf - ri
Suppose that an object moves from point A(ri) to point B(rf) in the x-y plane:
The Displacement
Vector is:  Δr = rf - ri
If this happens in in a time
Δt = tf - ti
The Average Velocity is:
vavg  (Δr/Δt)
Obviously, this in the same
direction as the displacement.
It is independent of the path
between A & B

4.  The instantaneous velocity v ≡ time derivative of displacement r
As Δt gets smaller & smaller, clearly, A & B get closer & closer together. Just as in 1d, we define The Instantaneous Velocity
 Velocity at Any Instant of Time.
 average velocity over an infinitesimally short time
Mathematically, the instantaneous velocity is:
v  lim∆t  0 [(∆r)/(∆t)] ≡ (dr/dt)
lim ∆t  0  ratio (∆r)/(∆t) for smaller & smaller ∆t. Mathematicians call this a derivative.
 The instantaneous velocity
v ≡ time derivative of displacement r

5. Instantaneous Velocity v ≡ (dr/dt).
The magnitude |v| of vector v ≡ speed. As motion progresses, the speed & direction of v can both change. For an object moving from A (vi) to B (vf) in the x-y plane:
Velocity Change  Δv = vf - vi
This happens in time Δt = tf - ti
Average Acceleration
aavg  (Δv/Δt)
As both the speed &
direction of v change,
over an arbitrary path

6. a ≡ time derivative of velocity v
As Δt gets smaller & smaller, clearly, A & B get closer & closer together. Just as in 1d, we define The Instantaneous Acceleration
 Acceleration at Any Instant of Time.
 average acceleration over infinitesimally short time
Mathematically, instantaneous acceleration is:
a ≡ lim∆t  0 [(∆v)/(∆t)] ≡ (dv/dt)
lim ∆t  0  ratio (∆v)/(∆t) for smaller & smaller ∆t. Mathematicians call this a derivative.
 The instantaneous acceleration
a ≡ time derivative of velocity v

7. 2d Motion, Constant Acceleration
It can be shown that:
Motion in the x-y plane can
be treated as 2 independent
motions in the x & y directions.
 So, motion in the x direction doesn’t affect the y motion & motion in the y direction doesn’t affect the x motion.

8. Suppose an object moves from A (ri,vi), to B (rf,vf), in the x-y plane
Suppose an object moves from A (ri,vi), to B (rf,vf), in the x-y plane. So, it’s position clearly changes with time:
The acceleration a is constant, so, as in 1d, we can write (vectors!):  rf = ri + vit + (½)at2
If its velocity also changes with time, it is accelerating. If it’s acceleration a is constant, so, as in 1d, we can write (vectors!):  vf = vi + at

9.  rf = ri + vit + (½)at2, vf = vi + at
Summary
For 2d motion when the acceleration a is constant, we have:
 rf = ri + vit + (½)at2, vf = vi + at
For the Horizontal Part of the Motion we have:
xf = xi + vxit + (½)axt2, vxf = vxi + axt
For the Vertical Part of the Motion we have:
yf = yyi + vyit + (½)ayt2, vyf = vyi + ayt

10. Projectile Motion

11. A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.
Figure Caption: This strobe photograph of a ball making a series of bounces shows the characteristic “parabolic” path of projectile motion.

12. Projectile Motion
Projectile Motion  Motion of an object that is projected into the air at an angle.
Near the Earth’s surface, the acceleration a on the projectile is downward and equal to
a = g = 9.8 m/s2
Goal: Describe the motion after it starts.
Galileo: Analyzed horizontal & vertical components of motion separately.
Today: Displacement D & velocity v are vectors
 Components of the motion can
be treated separately

13. Take down positive for a while! (vyf) 2 = (vyi)2 + 2g (yf - yi)
Equations Needed
One dimensional, constant acceleration equations for x & y separately!
x part: Acceleration ax = 0!
y part: Acceleration ay =  g!
(+ if down is positive - if down is negative).
Take down positive for a while!
The initial x & y components of the velocity: vxi & vyi.
x motion: vxf = vxi = constant. xf = xi + vxi t
y motion: vyf = vyf + gt, yf = yi + vyi t + (½)g t2
(vyf) 2 = (vyi)2 + 2g (yf - yi)

14. Projectile Motion  vyf = gt, yf = (½)g t2  vxf = vxi , xf = vxit
Simplest Example: A ball rolls across the table, to the edge & falls off edge to floor. It leaves the table at time t = 0. Analyze the x & y parts of the motion
separately.
y Part of the Motion: Down is positive. The origin is at the table top: yi = 0. Initially, there is no y component of velocity: vyi = 0
 vyf = gt, yf = (½)g t2
x Part of the Motion: The origin is at the table top: xi = 0. There is NO x component of acceleration(!)  ax = 0. Initially, the x component of velocity is: vxi
 vxf = vxi , xf = vxit

15. The Ball Rolls Across the Table & Falls Off
t = 0 here 
Can be understood by analyzing the horizontal & vertical motions separately. Take down as positive. The initial velocity has an x component ONLY!
That is vyi = 0.
vxi
At any point, v has both x & y components. The kinematic equations tell us that, at time t,
vx = vxi, vy = gt
xf = vxit
yf = (½)gt2
Figure Caption: Projectile motion of a small ball projected horizontally. The dashed black line represents the path of the object. The velocity vector at each point is in the direction of motion and thus is tangent to the path. The velocity vectors are green arrows, and velocity components are dashed. (A vertically falling object starting at the same point is shown at the left for comparison; vy is the same for the falling object and the projectile.)

16. vxf = vxi , vyf = gt xf = vxit , yf = (½)gt2 Summary 2 Components:
A ball rolling across the table & falling.
The vector velocity v has
2 Components:
vxf = vxi , vyf = gt
The vector displacement D also has
2 components:
xf = vxit , yf = (½)gt2

17. The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.
The photo shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction.
It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.
Figure Caption: Multiple-exposure photograph showing positions of two balls at equal time intervals. One ball was dropped from rest at the same time the other was projected horizontally outward. The vertical position of each ball is seen to be the same at each instant.

18. The PHYSICS of Projectile Motion
The vertical (y) part of the motion:
vyf = gt , yf = (½)g t2
Is the SAME as free fall!!
The horizontal (x) part of the motion:
vxf = vxi, xf = vxit
Is motion at constant velocity!!
 An object projected horizontally will reach the ground at the same time as an object dropped vertically from the same point!
(the x & y motions are independent)

19. General Case: An object is launched at initial angle θ0 with the horizontal. The analysis is similar to before, except the initial velocity has a vertical component vy0  0. Let up be positive now!
vx0 = v0cosθ0
vy0 = v0sinθ0
The acceleration = g
downward for the
entire motion!
Figure Caption: Path of a projectile fired with initial velocity v0 at angle θ0 to the horizontal. Path is shown dashed in black, the velocity vectors are green arrows, and velocity components are dashed. The acceleration a = dv/dt is downward. That is, a = g = -gj where j is the unit vector in the positive y direction.
The Parabolic shape of the path is
real (neglecting air resistance!)

20. Projectile Motion Diagram

21. vxi = vicosθi, vyi = visinθi NO ACCELERATION IN THE x DIRECTION!
General Case: Take y positive upward & origin at the point where it is shot: xi = yi = 0
vxi = vicosθi, vyi = visinθi
Horizontal motion:
NO ACCELERATION IN THE x DIRECTION!
vxf = vxi , xf = vxi t
Vertical motion:
vyf = vyi - gt , yf = vyi t - (½)gt2
(vyf) 2 = (vyi)2 - 2gyf
If y is positive downward, the - signs become + signs.
ax = 0, ay = -g = -9.8 m/s2

22. Summary: Projectile Motion
Projectile motion is motion with constant acceleration in two dimensions, where
The acceleration is g and is down.

23. Projectile Motion: More Discussion
As we already said, the y-component of the velocity is zero at the maximum height of the trajectory
The acceleration stays the same throughout the trajectory

24. Range & Maximum Height
When analyzing projectile motion, two characteristics are of special interest:
The Range, R = the horizontal distance of the projectile
The Maximum Height h = the highest point.

25. (vyf) 2 = (vyi)2 + 2g (yf - yi)
Maximum Height
The Maximum Height h of a projectile is the height at which the y component of velocity vy = 0.
It can be obtained from the kinematic equation:
(vyf) 2 = (vyi)2 + 2g (yf - yi)
Assuming yi = 0, setting vyf = 0 & putting yf = h:
0 = (vyi)2 + 2gh
So: h = [(vyi)2/(2g)]
Using vyi = vi sinθi gives:
Note that this is valid only for symmetric motion

26. Range
The Range of a Projectile, R = xmax, is the maximum horizontal distance from the starting point.
It can be obtained from the kinematic equation
R = xmax = vxitmax (1)
with tmax = the time for it to hit the ground.
The time to hit tmax can be obtained by setting yf = 0 in the kinematic equation yf = vyi t - (½)gt2. This gives
tmax = (2vyi/g) (2)
Putting (2) into (1) gives
R = (2vxivyi/g) (3)
In (3), using vxi = vicosθi & vyi = visinθi along with the trig identity 2 sinθicosθi = sin(2θi) gives the final result:
Note that this is valid only for a symmetric trajectory

27. Ranges for Fixed vi & Changing θi

28. Range The Range is: The maximum range R occurs at qi = 45o,
Using differential calculus, setting (dR/dqi) = 0 & solving for qi gives the result that
The maximum range R occurs at qi = 45o,
independent of the initial velocity vi.
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

29. Non-Symmetric Projectile Motion
Follow the general rules for projectile motion
One way is to 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 & solve the necessary equations