Chapter 3
Kinematics in Two Dimensions:
Projectile motion
Displacement, Velocity, and Acceleration

ro = initial position

r = final position
  
Δr = r − ro = displacement
!
In 2D, Δr is,!in general,
not
!
aligned with r and r0 , and not
in the direction of the path of
the motion of the body (e.g. the car).
Displacement, Velocity, and Acceleration
Average velocity is the
displacement divided by
the elapsed time.
 

 r − ro Δr
v=
=
t − to Δt
Displacement, Velocity, and Acceleration
The instantaneous velocity indicates how fast
the car moves and the direction of motion at each
instant of time.
!
!
Δr
v = lim
Δt→0 Δt
path of the car
!
v is in the direction
of the path of the car
Displacement, Velocity, and Acceleration
DEFINITIONS OF
AVERAGE
and
INSTANTANEOUS
ACCELERATION
ACCELERATION
 

 v − v o Δv
a=
=
t − to
Δt
!
!
Δv
a = lim
Δt→0 Δt

Δv

v

vo
!
!
Δv and a can point in very
different directions from
!
!
v 0 and v
Deriving the constant acceleration equations for 2-dimensions
 
 
 r − ro
 v − vo
v=
a=
t − to
t − to

Assume t0 = 0 and r0 = 0

 
 
 v − v0
 r
  
∴ v = ⇒ r = vt and a =
⇒ v = v 0 + at
t
t
 

 
Assume constant acceleration: a = a and v = 12 ( v 0 + v )
 
  

∴ r = 12 ( v 0 + v ) t and v = v 0 + at
Writing the vectors in component notation and substituting in the above:




r = xx̂ + yŷ v 0 = v0 x x̂ + v0 y ŷ v = vx x̂ + vy ŷ a = ax x̂ + ay ŷ
∴ xx̂ + yŷ = 12 ( v0 x x̂ + v0 y ŷ + vx x̂ + vy ŷ ) t and vx x̂ + vy ŷ = v0 x x̂ + v0 y ŷ + ( ax x̂ + ay ŷ ) t
Rearranging in terms of common factors in x̂ and ŷ:
$% x − 12 ( v0 x + vx ) t &' x̂ + $% y − 12 ( v0 y + vy ) t &' ŷ = 0 and [ vx − v0 x − ax t ] x̂ + $%vy − v0 y − ay t &' ŷ = 0
Since x̂ is perpendicular to ŷ, the equations can only be satisfied if each [] = 0
∴ x = 12 ( v0 x + vx ) t
y = 12 ( v0 y + vy ) t
vx = v0 x + ax t
vy = v0 y + ay t
and the other constant acceleration equations can be derived from these as done before
Equations of Kinematics in Two Dimensions
v x = vox + a x t
x = vox t + a x t
1
2
x=
2
2
x
1
2
(vox + vx )t
2
ox
v = v + 2a x x
Equations of Kinematics in Two Dimensions
v y = voy + a y t
y = voy t + a y t
1
2
2
y = 12 ( voy + vy ) t
2
y
2
oy
v = v + 2a y y
Equations of Kinematics in Two Dimensions
The x part of the motion occurs exactly as it would if the
y part did not occur at all, and vice versa.
Equations of Kinematics in Two Dimensions
Example: A Moving Spacecraft
In the x direction, the spacecraft has an initial velocity component
of +22 m/s and an acceleration of +24 m/s2. In the y direction, the
analogous quantities are +14 m/s and an acceleration of +12 m/s2.
Find (a) x and vx, (b) y and vy, and (c) the magnitude and direction
of the final velocity of the spacecraft at time 7.0 s.
Equations of Kinematics in Two Dimensions
Reasoning Strategy
1. Make a drawing.
2. Decide which directions are to be called positive (+) and
negative (-).
3. Write down the values that are given for any of the five
kinematic variables associated with each direction.
4. Verify that the information contains values for at least three
of the kinematic variables. Do this for x and y. Select the
appropriate equation.
5. When the motion is divided into segments, remember that
the final velocity of one segment is the initial velocity for the next.
6. Keep in mind that there may be two possible answers to a
kinematics problem.
Equations of Kinematics in Two Dimensions
Example: A Moving Spacecraft
In the x direction, the spacecraft has an initial velocity component
of +22 m/s and an acceleration of +24 m/s2. In the y direction, the
analogous quantities are +14 m/s and an acceleration of +12 m/s2.
Find (a) x and vx, (b) y and vy, and (c) the magnitude and direction
of the final velocity of the spacecraft at time 7.0 s.
x
ax
vx
vox
t
?
+24 m/s2
?
+22 m/s
7.0 s
y
ay
vy
voy
t
?
+12 m/s2
?
+14 m/s
7.0 s
Equations of Kinematics in Two Dimensions
x
ax
vx
vox
t
?
+24 m/s2
?
+22 m/s
7.0 s
x = vox t + ax t
1
2
2
= ( 22 m s) ( 7.0 s) +
2
1
2
(24 m s ) ( 7.0 s)
2
) ( 7.0 s) = +190 m s
2
= +740 m
vx = vox + ax t
= ( 22 m s) + ( 24 m s
Equations of Kinematics in Two Dimensions
y
ay
vy
voy
t
?
+12 m/s2
?
+14 m/s
7.0 s
y = voy t + ay t
1
2
2
= (14 m s) ( 7.0 s) + (12 m s
1
2
2
) ( 7.0 s)
2
= +390 m
vy = voy + ay t
= (14 m s) + (12 m s
2
) ( 7.0 s) = +98m s
Final position and velocity vectors in component form:
!
r = (740 m)x̂ + (390 m)ŷ
!
v = (190 m s)x̂ + (98m s)ŷ
Equations of Kinematics in Two Dimensions
Equations of Kinematics in Two Dimensions
v
v y = 98 m s
θ
v x = 190 m s
v=
(190 m s)
θ = tan
−1
2
2
+ ( 98m s) = 210 m s
(98 190) = 27

Projectile Motion
Under the influence of gravity alone, an object near the
surface of the Earth will accelerate downwards at 9.80m/s2.
a y = −9.80 m s
2
ax = 0
v x = vox = constant
Projectile Motion
Conceptual Example:
I Shot a Bullet into the Air...
Suppose you are driving a convertible with the top down.
The car is moving to the right at constant velocity. You point
a rifle straight up into the air and fire it. In the absence of air
resistance, where would the bullet land – behind you, ahead
of you, or in the barrel of the rifle?
Projectile Motion
Example: A Falling Care Package
The airplane is moving horizontally with a constant velocity of
+115 m/s at an altitude of 1050 m. Determine the time required
for the care package to hit the ground.
Projectile Motion
y
ay
-1050 m -9.80 m/s2
vy
voy
t
0 m/s
?
Projectile Motion
y
ay
vy
-1050 m -9.80 m/s2
y = voy t + a y t
1
2
2
voy
t
0 m/s
?
y = a yt
1
2
2
2y
2(− 1050 m )
t=
=
=
14
.
6
s
2
ay
− 9.80 m s
Projectile Motion
Example: The Velocity of the Care Package
What are the magnitude and direction of the final velocity of
the care package?
Projectile Motion
y
ay
-1050 m -9.80 m/s2
vy
voy
t
?
0 m/s
14.6 s
Projectile Motion
y
ay
-1050 m -9.80 m/s2
vy
voy
t
?
0 m/s
14.6 s
vy = voy + ay t = 0 + (−9.80 m s
= −143m s
2
) (14.6 s)
Find the final velocity of the package.

v = vx x̂ + vy ŷ
= (115 m/s)x̂ + (−143 m/s)ŷ
= - 143 m/s
v2 = vx2 + vy2
= (115 m/s)2 + (143 m/s)2
= 33700 (m/s)2
à  v = 184 m/s
θ = tan-1 (143/115) = 51.2o