2D Equations of Motion constant acceleration Summary Table 41 from Walker

1
2D Equations of Motion(constant acceleration)
SummaryTable 4-1 from Walker
These equations will be used for all of the
results in Ch. 4!
2
Zero Launch Angle
x0 0, y0 h v0x v0, v0y 0
3
Landing site for Zero Launch Angle

• Where does a projectile land if it is launched
  horizontally with a speed v0 from a height h?
• 
  (solution on the board)

h
x ?
4
Zero Launch Angle Example 1

• Problem 15 p. 99. A sparrow flying horizontally
  with a speed of 1.80 m/s folds its wings and
  begins to drop in freefall.
• (a) How far does the sparrow fall after traveling
  a horizontal distance of 0.500 m?
• (b) If the sparrows initial speed is increased,
  does the distance of fall increase, decrease, or
  stay the same?

5
General launch angle
x0 0, y0 h v0x v0 cos ? v0y v0 sin ?
6
Projectile Motion

• Our reference frame

7
Projectile Motion General launch angle

• Reference frame chosen
• y is vertical with upward positive
• Acceleration components
• ay -g and ax 0
• Initial velocity components
• vxi vi cos q and vyi vi sin q

• The velocity components for the projectile at any
  time t are
• vxf vxi vi cos qi constant
• vyf vyi ayt vyi g t vi sin qi g t

8
Projectile Motion Equations
- g
9
General launch angle equations
x0 0, y0 0 v0x v0 cos ?, v0y v0 sin ?
using
10
Projectile Motion Position(Verifying the
Parabolic Trajectory)

• Position
• x0 y0 0
• xf vxi t (vi cos q) t 1
• yf vyi t 1/2ay t2 (vi sin q)t - 1/2 gt2
  2
• Solve 1 for t and use that in 2 gives us
• This is in the form of y ax bx2 which is the
  standard form of a parabola that passes through
  the origin.

y
x
a
b
11
Analyzing Projectile Motion

• Consider the motion as the superposition of the
  motions in the x- and y-directions
• The x-direction has constant velocity
• ax 0
• The y-direction is free fall
• ay -g

12
Projectile Motion Implications

• The y-component of the velocity is zero at the
  maximum height of the trajectory
• The acceleration stays the same throughout the
  trajectory
• The x-component of the velocity is constant

13
The time of a projectile flight (How long a
projectile travels before landing?)

• Recall y (v0 sin ?) t - 1/2gt2 (for
  arbitrary launching angle)
• yi yf 0 (same initial and final elevation)
• 0 (v0 sin ?) t - 1/2gt2
• (v0 sin ?) t 1/2gt2
• v0 sin ? 1/2gt

14
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

15
Maximum height?

• At the maximum height, vy 0!
• So, vyi g t vi sin qi g t 0 and,
• Is the time when the particle is at the maximum
  height. To find this height, use the relation
  for vertical position with this time yf (vi
  sin q)t - 1/2 gt2
• Gives h (vi sin q)(vi sin q/g) 1/2 g (vi sin
  q/g)2

16
Height of a Projectile, equation

• So, the maximum height of the projectile can be
  found in terms of the initial velocity vector
• What are the consequences?
• Increase h by making vi or qi bigger or g smaller

17
Range of a Projectile, equation

• Maximum range is TWICE the horizontal position at
  the time of maximum height. (For the geometry of
  our case same initial and final elevation)
• In general x (vi cos qi) t so,
• Now, 2 sin q cos q sin 2q, so

18
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

19
More About the Range of a Projectile
No air resistance!
20
Range of a Projectile, final

• The maximum range occurs at qi 45o
  sin(245o)1
• Complementary angles will produce the same range
• recall that angles t and t' are
  complementary values if tt' 90 o,
• sin (2t') sin2(90
  - t) sin(180-2t) sin(2t)
• e.g. sin(230o) sin(260o)
  sin(180o-60o) sin60o
• The maximum height will be different for the two
  angles
• The times of the flight will be different for
  the two angles

21
Projectiles with Air Resistance
22
Symmetry in Projectile Motion

• If launching and landing heights are the same (e.
  g. y 0)
• The time when the projectile lands 1/2 of the
  time to reach its highest point
• When the projectile lands its speed the speed
  of launching (velocities are different!)
• The angle of velocity above the horizontal on the
  way up The angle of velocity below the
  horizontal on the way down

23
Projectile Motion Problem Solving Hints

• Conceptualize
• Establish the mental representation of the
  projectile moving along its trajectory
• Analyze
• Resolve the initial velocity into x and y
  components
• Remember to treat the horizontal motion
  independently from the vertical motion
• Treat the horizontal motion using constant
  velocity techniques
• Analyze the vertical motion using constant
  acceleration techniques
• Remember that both directions share the same time
• Finalize
• Check your results

24
Conceptual questions (p.97 98)

• Q. 4. Is it possible for the velocity of a
  projectile to be at right angles to its
  acceleration? If so, give an example.
• Q. 8. A projectile is launched from a level
  surface with an initial velocity of
  What is the velocity of the
  projectile just before it lands?
• Q. 9. A projectile is launched with an initial
  velocity of
  What is the velocity of the projectile when it
  reaches its highest point?

25
Conceptual questions (p.97 98)

• Q. 14. Three projectiles (a, b, and c) are
  launched with different initial speeds so that
  they reach the same maximum height, as shown in
  Figure. List the projectiles in order of
  increasing (a) initial speed and (b) time of
  flight.

26
Zero Launch Angle Example 2

• Problem 21 p. 100. A basketball is thrown
  horizontally with an initial speed of 4.20 m/s. A
  straight line drawn from the release point to the
  landing point makes an angle of 30.0 with the
  horizontal. What was the release height?

27
Non-Symmetric Projectile Motion

• Follow the general rules for projectile motion
• or
• Break the y-direction into parts
• up and down or
• symmetrical back to initial height and then the
  rest of the height
• follow the general rules for projectile motion

28
Projectile Motion Example 1

• A firefighter a distance 50.0 m from a burning
  building directs a stream of water from a fire
  hose at angle ?i 30 above the horizontal. If
  the initial speed of the stream is 40.0 m/s, at
  what height does the water strike the building?

29
Projectile Motion Example 1, cont

• Water hits the building when

• At this time, the height (y) of the water is

30
Projectile Motion Example 2

• A ball is tossed from an upper-story window of a
  building. The ball is given an initial velocity
  of 8.00 m/s at an angle of 20.0 below the
  horizontal. It strikes the ground 3.00 s later.
• How far horizontally from the base of the
  building does the ball strike the ground?
• Find the height from which the ball was thrown.
• How long does it take the ball to reach a point
  10.0 m below the level of launching?

y
vi 8.00 m/s t 3.00 s
x
20.0º
h ?
x ?
31
Projectile Motion Example 2, cont.
y
x

• (a)

20.0º
vi 8.00 m/s

• (b)

52.3 m below starting point
32
Projectile Motion Example

• (c) Now take final point 10 m below window

Physically meaningful answer