Topic 9: Motion in fields
9.1 Projectile motion
9.1.1 State the independence of the vertical and
the horizontal components of velocity for a
projectile in a uniform field.
9.1.2 Describe and sketch the trajectory of
projectile motion as parabolic in the absence
of air resistance.
9.1.3 Describe qualitatively the effect of air
resistance on the trajectory of a projectile.
9.1.4 Solve problems on projectile motion.
Topic 9: Motion in fields
9.1 Projectile motion
State the independence of the vertical and the
horizontal components of velocity for a
projectile in a uniform field.
A projectile is an object that has been given an
initial velocity by some sort of short-lived
force, and then moves through the air under the
influence of gravity.
Baseballs, stones, or bullets are all examples
of projectiles.
You know that all objects moving through air feel
an air resistance (recall sticking your hand out
of the window of a moving car).
FYI
We will ignore
air resistance in
the discussion
that follows…
Topic 9: Motion in fields
9.1 Projectile motion
ay = -g
Speeding up in -y
dir.
ay = -g
Slowing down in +y
dir.
State the independence of the vertical and the
horizontal components of velocity for a
projectile in a uniform field.
Regardless of the air resistance, the vertical
and the horizontal components of velocity of an
object in projectile motion are independent.
Constant speed in +x dir. ax = 0
Topic 9: Motion in fields
9.1 Projectile motion
Describe and sketch the trajectory of projectile
motion as parabolic in the absence of air
resistance.
The trajectory of a projectile in the absence of
air is parabolic. Know this!
Topic 9: Motion in fields
9.1 Projectile motion
Describe qualitatively the effect of air
resistance on the trajectory of a projectile.
If there is air resistance, it is proportional to
the square of the velocity. Thus, when the ball
moves fast its deceleration is greater than when
it moves slow.
SKETCH POINTS
Peak to left of
original one.
Pre-peak
distance more
than post-peak.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
Recall the kinematic equations from Topic 2:
s = ut + (1/2)at2 Displacement
v = u + at
Velocity
kinematic
equations
a is constant
Since we worked only in 1D at the time, we didn’t
have to distinguish between x and y in these
equations.
Now we appropriately modify the above to meet our
new requirements of simultaneous equations:
∆x = uxt + (1/2)axt2
vx = ux + axt
∆y = uyt + (1/2)ay
vy = uy + ayt
t2
kinematic
equations
ax and ay are constant
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
0
∆x = uxt + (1/2)axt2
0
vx = ux + axt
∆y = uyt + (1/2)ayt2
vy = uy + ayt
kinematic
equations
ax and ay are constant
PRACTICE: Show that the reduced equations for
projectile motion are
∆x = uxt
∆y = uyt - 5t2
vx = ux
vy = uy - 10t
reduced equations
of projectile
motion
SOLUTION:
ax = 0 in the absence of air resistance.
ay = -10 in the absence of air resistance.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
∆x = uxt
∆y = uyt - 5t2
vx = ux
vy = uy - 10t
reduced equations
of projectile
motion
EXAMPLE: Use the reduced equations above to prove
that projectile motion (in the absence of air
resistance) is parabolic.
SOLUTION: Just solve for t in the first equation
and substitute it into the second equation.
∆x = uxt becomes t = ∆x/ux so that t2 = ∆x2/ux2.
Then ∆y = uyt - 5t2, or
∆y = (uy/ux)∆x – (5/ux2)∆x2.
FYI
The equation of a parabola is y = Ax + Bx2.
In this case, A = uy/ux and B = -5/ux2.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
∆x = uxt
∆y = uyt - 5t2
vx = ux
vy = uy - 10t
reduced equations
of projectile
motion
PRACTICE: A cannon fires a projectile with a
muzzle velocity of 56 ms-1 at an angle of
inclination of 15º.
(a) What are ux and uy?
SOLUTION: Make a
velocity triangle.
uy = u sin 
 = 15º
uy = 56 sin 15º
ux = u cos 
uy = 15 m s-1.
ux = 56 cos 15º
ux = 54 m s-1
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
∆x = uxt
∆y = uyt - 5t2
vx = ux
vy = uy - 10t
reduced equations
of projectile
motion
PRACTICE: A cannon fires a projectile with a
muzzle velocity of 56 ms-1 at an angle of
inclination of 15º.
(b) What are the tailored equations of motion?
(c) When will the ball reach its maximum height?
SOLUTION: (b) Just substitute ux = 54 and uy = 15:
∆x = 54t
∆y = 15t - 5t2
tailored equations
for this particular
projectile
vx = 54
vy = 15 - 10t
(c) At the maximum height, vy = 0. Why? Thus
vy = 15 - 10t becomes 0 = 15 - 10t so that
10t = 15
t = 1.5 s.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
∆x = uxt
∆y = uyt - 5t2
vx = ux
vy = uy - 10t
reduced equations
of projectile
motion
PRACTICE: A cannon fires a projectile with a
muzzle velocity of 56 ms-1 at an angle of
inclination of 15º.
(d) How far from the muzzle will the ball be when
it reaches the height of the muzzle at the end of
its trajectory?
SOLUTION:
From symmetry tup = tdown = 1.5 s so t = 3.0 s.
Thus
∆x = 54t
∆x = 54(3.0)
∆x = 160 m.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
∆x = uxt
∆y = uyt - 5t2
vx = ux
vy = uy - 10t
reduced equations
of projectile
motion
PRACTICE: A cannon fires a projectile with a
muzzle velocity of 56 ms-1 at an angle of
inclination of 15º.
(e) Sketch the following graphs:
a vs. t,
vx vs. t,
and vy vs. t:
SOLUTION:
ay
The only acceleration is
-10
g in the –y-direction.
vx
vx = 54, a constant. Thus
54
it does not change over time.
vy
vy = 15 - 10t Thus it is
15
linear with a negative gradient
1.5
and it crosses the time axis at 1.5 s.
t
t
t
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
The acceleration is ALWAYS g for projectile
motion-since it is caused by Earth and its field.
At the maximum height the projectile switches
from upward to downward motion. vy = 0 at switch.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
The flight time is
limited by the y
motion.
The maximum height
is limited by the y
motion.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
ax = 0.
ay = -10 ms-2.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
∆y = uyt - 5t2
-33 = 0t - 5t2
-33 = -5t2
(33/5) = t2
Fall time limited by y-equations:
t = 2.6 s.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
∆x = uxt
∆x = 18(2.6)
Use x-equations and t = 2.6 s:
∆x = 15 m.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
18
vx = ux
vx = 18.

vy = uy – 10t
vy = 0 – 10t
vy = –10(2.6) = -26.
tan  = 26/18
 = tan-1(26/18) = 55º.
26
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
The horizontal component of velocity is vx = ux
which is CONSTANT.
The vertical component of velocity is
vy = uy – 10t which is INCREASING (negatively).
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
 ∆EK + ∆EP = 0
∆EK = -∆EP
∆EK = -mg∆y
EK0 = (1/2)mu2
∆EK = -(0.44)(9.8)(-32) = +138 J = EK – EK0
EK =
+138
+ (1/2)(0.44)(222) = 240 J.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
If 34% of the energy is
consumed, 76% remains.
0.76(240) = 180 J
(1/2)(0.44)v2 = 180 J
v = 29 ms-1.
(1/2)mvf2 - (1/2)mv2 = -∆EP
mvf2 = mv2 + -2mg(0-H)
Topic 9: Motion in fields
vf2 = v2 + 2gH
9.1 Projectile motion
Solve problems on projectile motion.
Use ∆EK + ∆EP = 0.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
uy = u sin 
uy = 28 sin 30º
ux = u cos 
ux = 28 cos 30º
ux = 24 m s-1.
uy = 14 m s-1.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
∆x = uxt
16 = 24t
t = 16/24 = 0.67
The time to the wall is found from ∆x…
∆y = uyt – 5t2
∆y = 14t – 5t2
∆y = 14(0.67) – 5(0.67)2 = 7.1 m.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
0.5s
0.0s
4 m
ux = ∆x/∆t = (4-0)/(0.5-0.0) = 8 ms-1.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
0.5s
11 m
0.0s
4 m
uy = ∆y/∆t = (11-0)/(0.5-0.0) = 22 ms-1.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
2.5s 3.0s
2.0s 30 m
1.5s
1.0s
0.5s
11 m
0.0s
-1(30/24) = 51º

=
tan

4 m
24 m
D2 = 242 + 302 so that D = 38 m ,@  = 51º.
Topic 9: Motion in fields
9.1 Projectile motion
Solve problems on projectile motion.
New peak
below and
left.
Pre-peak
greater
than
postpeak.