1) The document discusses curvilinear motion, which refers to motion along a curved path rather than a straight line. It defines position, velocity, and acceleration vectors for particles undergoing curvilinear motion. 2) As an example, it examines the motion of projectiles, noting that the horizontal and vertical components of motion can be treated independently as rectangular components and integrated separately. 3) It then works through an example problem calculating the horizontal distance and maximum height for a projectile fired from the edge of a cliff.

1. Engineering DYNAMICS
Lecture
7 Kinematics of Particles:
Projectile Motion

2. The softball and the car both undergo
curvilinear motion.
• A particle moving along a curve other than a
straight line is in curvilinear motion.
Curvilinear Motion: Position, Velocity & Acceleration

3. • The position vector of a particle at time t is defined by a vector between
origin O of a fixed reference frame and the position occupied by particle.
• Consider a particle which occupies position P defined by at time t
and P’ defined by at t + Dt,
r

r

Curvilinear Motion: Position, Velocity & Acceleration

4. 0
lim
t
s ds
v
t dt
D 
D
 
D
Instantaneous velocity
(vector)
Instantaneous speed
(scalar)
0
lim
t
r dr
v
t dt
D 
D
 
D
Curvilinear Motion: Position, Velocity & Acceleration

5. 0
lim
t
v dv
a
t dt
D 
D
  
D
instantaneous acceleration (vector)
• Consider velocity of a particle at time t and velocity at t + Dt,
v

v


• In general, the acceleration vector is not tangent
to the particle path and velocity vector.
Curvilinear Motion: Position, Velocity & Acceleration

6. • When position vector of particle P is given by its
rectangular components,
k
z
j
y
i
x
r







• Velocity vector,
k
v
j
v
i
v
k
z
j
y
i
x
k
dt
dz
j
dt
dy
i
dt
dx
v
z
y
x






















• Acceleration vector,
k
a
j
a
i
a
k
z
j
y
i
x
k
dt
z
d
j
dt
y
d
i
dt
x
d
a
z
y
x
























 2
2
2
2
2
2
Rectangular Components of Velocity & Acceleration

7. Example 2/5

8. Example 2/5

9. • Rectangular components particularly effective
when component accelerations can be integrated
independently, e.g., motion of a projectile,
0
0 





 z
a
g
y
a
x
a z
y
x 





with initial conditions,
      0
,
,
0 0
0
0
0
0
0 


 z
y
x v
v
v
z
y
x
Integrating twice yields
𝑣𝑥 = 𝑣𝑥 0 𝑣𝑦 = 𝑣𝑦 0
− 𝑔𝑡 𝑣𝑧 = 0
𝑥 = 𝑣𝑥 0𝑡 𝑦 = 𝑣𝑦 0
𝑡 −
1
2
𝑔𝑡2
𝑧 = 0
• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly accelerated.
• Motion of projectile could be replaced by two
independent rectilinear motions.
Rectangular Components for Projectile motion

10. A projectile is fired from the edge
of a 150-m cliff with an initial
velocity of 180 m/s at an angle of
30°with the horizontal. Neglecting
air resistance, find (a) the horizontal
distance from the gun to the point
where the projectile strikes the
ground, (b) the greatest elevation
above the ground reached by the
projectile.
SOLUTION:
• Consider the vertical and horizontal motion
separately (they are independent)
• Apply equations of motion in y-direction
• Apply equations of motion in x-direction
• Determine time t for projectile to hit the
ground, use this to find the horizontal
distance
• Maximum elevation occurs when vy=0
Problem

11. SOLUTION:
Given: (v)o =180 m/s (y)o =150 m
(a)y = - 9.81 m/s2
(a)x = 0 m/s2
Vertical motion – uniformly accelerated:
Horizontal motion – uniformly accelerated:
Choose positive x to the right as shown

12. SOLUTION:
Horizontal distance
Projectile strikes the ground at:
Solving for t, we take the positive root
Maximum elevation occurs when vy=0
Substitute into equation (2) above
Substitute t into equation (4)
Maximum elevation above the ground =