1/17/2014
2-d Motion: Constant Acceleration
• Kinematic Equations of Motion (Vector Form)
Acceleration Vector (constant)
Velocity Vector (function of t)
Position Vector (function of t)
The velocity vector and position
vector are a function of the time t.
• Kinematic Equations of Motion (Component Form)
r
Warning! These
a = a x xˆ + a y yˆ
equations are only
valid if the
r
r r
acceleration is
v (t ) = v0 + at
constant.
r
r r
r2
1
r (t ) = r0 + v0t + 2 at
Velocity Vector at
time t = 0.
ax =
ay =
constant
v x (t ) = v x 0 + a x t
x(t ) = x0 + v x 0t + 12 a xt 2
constant
Warning! These
equations are only
valid if the
acceleration is
constant.
v y (t ) = v y 0 + a y t
y (t ) = y0 + v y 0t + 12 a y t 2
The components of the acceleration vector, ax and ay, are constants.
The components of the velocity vector at t = 0, vx0 and vy0, are constants.
r
v0 = v x 0 xˆ + v y 0 yˆ
r
r0 = x0 xˆ + y0 yˆ
Position Vector at
time t = 0.
2-d Motion: Constant Acceleration
The components of the position vector at t = 0, x0 and y0, are constants.
• Ancillary Equations
Valid at any time t
v (t ) − v = 2a x ( x(t ) − x0 )
2
x
The components of the acceleration vector, ax and ay, are constants.
2
xo
2
v y2 (t ) − v yo
= 2 a y ( y (t ) − y 0 )
The components of the velocity vector at t = 0, vx0 and vy0, are constants.
The components of the position vector at t = 0, x0 and y0, are constants.
University of Florida
PHY 2053
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University of Florida
Projectile Motion
PHY 2053
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Rules of Projectile Motion
The x- and y-directions of motion completely independent
The x-direction is uniform motion: ax = 0
The y-direction is free fall: ay = -g
The initial velocity v 0 can be broken down into its x- and ycomponents
v = v cos θ
v = v sin θ
•
Ox
x-direction--ax
O
O
Oy
O
=0
v xo = v o cos θo = v x = constant
x = vxot This is the only operative equation in the x-direction since there is
O
uniform velocity in that direction
• y-direction--free fall:
a = -g
–take the positive direction as upward
–uniformly accelerated motion, so the motion equations all hold
v =
• Velocity at any time
Example: Projectile Motion
In this case, ax= 0 and ay= -g, vx0 = v0cosθ
θ, vy0 = v0sinθ
θ, x0 = 0, y0 = 0.
v x (t ) = v0 cos θ
x(t ) = (v0 cos θ )t
v y (t ) = v0 sin θ − gt
v0
y (t ) = (v0 sin θ )t − gt
θ
2
v x (t ) = v0 cos θ
x(t ) = (v0 cos θ )t
y-axis
y (t ) = (v0 sin θ )t − gt
The time, tmax, that the projective reaches its maximum
height occurs when vy(tmax) = 0. Hence,
v0 sin θ
(v0 sin θ ) 2
tmax =
2g
g
H = y (tmax ) =
2g
For a fixed v the largest
R
• Range R (maximum horizontal
distance
traveled)
The time, tf, that it takes the projective reach the
ground occurs when y(tf) = 0. Hence,
2v sin θ
The time, tf, that it takes the projective reachthe ground
occurs when y(tf) = 0. Hence,
2v sin θ
tf =
0
occurs when θ = 45o!
0 = y (t f ) = (v0 sin θ )t f − 12 gt 2f
0
g
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0
g
2
0
2
0
PHY 2053
tf =
2v sin θ cos θ v02 sin 2θ
R = x(t f ) = (v0 cos θ )t f =
=
g
g
2v sin θ cos θ v sin 2θ
R = x(t f ) = (v0 cos θ )t f =
=
g
g
2
0
θ
2
• Range R (maximum horizontal distance traveled)
0 = y (t f ) = (v0 sin θ )t f − 12 gt 2f
v0
x-axis
1
2
The time, tmax, that the projective reaches its maximum
height occurs when vy(tmax) = 0. Hence,
v0 sin θ
(v0 sin θ ) 2
H = y (tmax ) =
vx
v y (t ) = v0 sin θ − gt
• Maximum Height H
g
vy
In this case, ax= 0 and ay= -g, vx0 = v0cosθ
θ, vy0 = v0sinθ
θ, x0 = 0, y0 = 0.
• Maximum Height H
tmax =
θ = tan−1
• Near the Surface of the Earth (h = 0)
x-axis
1
2
and
Example: Projectile Motion
y-axis
• Near the Surface of the Earth (h = 0)
v x2 + v y2
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University of Florida
PHY 2053
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1/17/2014
Some Special Cases of Projectile Motion
Problem-Solving Strategy
Select a coordinate system and sketch the path
of the projectile--Include initial and final
positions, velocities, and accelerations
Resolve the initial velocity into x- and ycomponents
Treat the horizontal and vertical motions
independently
• Horizontal motion: Use techniques for problems
with constant velocity
Object may be fired
horizontally
The initial velocity is
all in the x-direction
• vo = vx and vy = 0
All the general rules of
projectile motion
apply
• Vertical motion: Use techniques for problems with
constant acceleration
Exam 1 Fall 2012: Problem 11
Exam 1 Spring 2012: Problem 12
v
• Near the surface of the Earth a projectile
θ
d
is fired from the top of a building at a
height h above the ground at an angle θ
h
building
relative to the horizontal and at a distance
d from the edge of the building as shown
ground
in the figure. If θ = 20o and d = 20 m, what
is the minimum initial speed, v0, of the
v02 sin 2θ
projectile such that it will make it off the d = R =
g
building and reach the ground? Ignoring
air resistance.
0
Answer: 17.5 m/s
% Right: 35%
Let th be the time the beanbag hits the ground.
dg
( 20m)(9.8m / s 2 )
v0 =
=
≈ 17.46m / s ≈ 17.5m / s
sin 2θ
sin(2 × 20o )
University of Florida
PHY 2053
Relative Position Vector
Position of car A relative
to car B is given by the
vector subtraction
equation
r
• rAE is the position of car A as
measured by E
r
• rBE is the position of car B
as measured by E
r
• rABis the position of car A as
measured by car B
r
r
r
rAB = rAE − rEB
• A beanbag is thrown horizontally from a dorm room
window a height h above the ground. It hits the ground a
horizontal distance d = h/2 from the dorm directly below
the window from which it was thrown. Ignoring air
resistance, find the direction of the beanbag's velocity
just before impact.
2h
y-axis
Answer: 76.0°below the horizontal
t h2 =
% Right: 22%
g
h
v y (t ) = − gt
v x (t ) = v0
d
d
x-axis
v0 =
2
1
θ
y
(
t
)
=
h
−
gt
th
x(t ) = v0t
2
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y (t h ) = 0 = h − 12 gth2
x(t h ) = d = v0th
University of Florida
tan θ =
v y (t h )
v x (t h )
gt h gt h2 2h
=
=
=4
v0
d
d
o
θ ≈ 76
=
PHY 2053
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Relative Velocity Notation
The pattern of subscripts can be useful in solving
relative velocity problems
Assume the following notation:
• E is an observer, stationary with respect to the
earth
• A and B are two moving cars
The rate of change of the displacements gives
the relationship for the velocities
r
r
r
v AB = v AE − v EB
2