Projectile Motion Notes
Vertical Projectile Motion
Vertical Projectile Motion
Case 1
Case 2
Case 3
Case 4
u
Object launchedObject launched
Object
Object
vertically
and dropped
vertically
and projected
u
returns to launch
returns todownwards
a
from rest
height
different height
u
Case 1: Vertical Projectile Motion
+
Set positive
direction as down
Use constant acceleration equations with
u = 0 a = 10ms-1
If time not involved can use energy approach
Ug  Ek
mgh = ½mv2
gh = ½v2
2gh = v2
Three points
Case 2: Vertical Projectile Motion
+
u
Set positive
direction as down
Use constant acceleration equations with
a = 10ms-1
If time not involved can use energy approach
Eki + Ug  Ekf
½mu2 + mgh = ½mv2
½u2 + gh = ½v2
u2 + 2gh = v2
Three points
Case 3: Vertical Projectile Motion
If time not involved can use
energy approach
Eki  Ekf + Ug
½mu2 = ½mv2 + mgh
½u2 = ½v2 + gh
u2 = v2 + 2gh
v=0
Use const accel equations
with a = -10ms-1
v
v
+
Set positive
direction as up
Six points
u
Speeds up are the
same as down at
each height
Time up = time down
or
Total time = 2 × time up
Case 4: Vertical Projectile Motion
If time not involved can use
energy approach
Eki  Ekf + Ug
½mu2 = ½mv2 + mgh
v=0
Use const accel equations
with a = -10ms-1
+
Set positive
direction as up
Six points
u
Speeds up are the
same as down at
each height
Displacements below
launch height will be
negative
Case 4: Vertical Projectile Motion
If time not involved can use
energy approach
Eki  Ekf + Ug
½mu2 = ½mv2 + mgh
v=0
Use const accel equations
with a = -10ms-1
+
Set positive
direction as up
Six points
u
Speeds up are the
same as down at
each height
Displacements below
launch height will be
negative
Vertical Projectile Motion
Worked Examples
Vertical Projectile Motion
Example 1
A stone is dropped from an 8.0m tower and falls to the
ground.
(a) How long will it take the stone to drop to the ground?
t=?
u=0
x = 8.0m
a = 10ms-2
x = ut + ½ at2
• In Maths when you determine the
square root of a number you should
8 = ½ × 10 × t2
always give the result as a  value.
2
This means there are two answers
8=5×t
+
• In Physics we normally just write out
2
1.6 = t
the magnitude of the square root
because the negative value is either
1.26491 = t
not relevant or it represents a
direction which is already obvious in
t  1.3 s
the problem. In this case a negative
time is not plausible.
Vertical Projectile Motion
Example 1
A stone is dropped from an 8.0m tower and falls to the
ground.
(b) What speed will the stone hit the ground?
v=?
u=0
x = 8.0m
a = 10ms-2
v2 = u2 + 2ax
Alternative – Energy Approach
2
v = 2 × 10 × 8
Ug just
Ekwrite out
•
In
In
Physics
we
normally
2
v2 = 160
the magnitude of
the square
root
mgh
= ½mv
because the negative
v = 12.6491
gh value
= ½vis2 either not
+
relevant or it represents a direction
-1
2
2gh =in vthe
v  13 ms
which is already obvious
problem.
In this case a negative
2 × 10 value
× 8 =means
v2 the
stone is moving upwards
160 =which
v2 is not
plausible.
12.6491 = v
v = 13ms-1
Vertical Projectile Motion
Example 2
A stone is dropped down a well and takes 2.5
seconds to hit the water. How deep is the well?
x=?
+
u=0
t = 2.5s
x = ut + ½ at2
x = ½ × 10 × 2.52
x = 31.25
x  31 s
a = 10ms-2
Vertical Projectile Motion
Example 3
A stone is thrown down at 2.0ms-1 from a 3.0m
tower and falls to the ground. What speed did the
ground hit the ground?
v=?
u=0
x = 10m
a = 10ms-2
Alternative – Energy Approach
+
v2 = u2 + 2ax
i+ U  E f
E
k
g
k
2
2
2.0ms-1
v = 2 + 2 × 10 × 3
½mu2 + mgh = ½mv2
½u2 + gh = ½v2
v2 = 64
2 + 2gh = v2
u
v = 8 ms-1
22 + 2 × 10 × 3 = v2
64 = v2
8=v
v = 8ms-1
Vertical Projectile Motion
Example 4
A cannon ball is fired upwards from the ground at a speed
of 30ms-1.
(a) How high will the cannon ball reach?
x=?
u = 30ms-1
v=0
a = –10ms-2
x
30ms-1
+
v2 = u2 + 2ax
0 = 302 + 2 × – 10 × x
0 = 900 + – 20 × x
– 900 = – 20 × x
45 = x
x = 45m
Vertical Projectile Motion
Example 4
A cannon ball is fired upwards from the ground at a speed
of 30ms-1.
(a) How high will the cannon ball reach?
x=?
u = 30ms-1
v=0
a = –10ms-2
v2
x
30ms-1
+
u2
= + 2ax
0 = 302 + 2 × – 10 × x
0 = 900 + – 20 × x
– 900 = – 20 × x
45 = x
x = 45m
Alternative – Energy Approach
EkUg
½mv2 = mgh
½v2 = gh
v2 = 2gh
302 = 2 × 10 × h
900 = 20h
45 = h
h = 45m
Vertical Projectile Motion
Example 4
A cannon ball is fired upwards from the ground at a speed
of 30ms-1.
(b) What will be the flight time of the cannon ball?
Find time to top of the path
t=?
u = 30ms-1 v = 0
a = –10ms-2
v = u + at
0 = 30 + – 10 × t
t t
– 30 = – 10 × t
3=t
+
Find total time
30ms-1
Total time = 2 × 3
= 6 seconds
Vertical Projectile Motion
Example 4
A cannon ball is fired upwards from the ground at a speed
of 30ms-1.
(b) What will be the flight time of the cannon ball?
Find
time
to
top
of
the
path
Alternative 1 -1
–10ms-1
t t==?? u u= =
30ms
v
=
0
a
=
-1
–
-2
30ms
x = 0 a = 10ms
vx==uut++at
½at2
–
00==30
30t++ ½10
× –×10t × t2
t t
– 030
==
30t– 10
+ – 5× ×t t2
0 = 5t
(6 – t)
3
t
+
t = 0, time
6
Find total
30ms-1
So flight
time
6 seconds
Total
time
= 2= ×
3
= 6 seconds
Vertical Projectile Motion
Example 4
A cannon ball is fired upwards from the ground at a speed
of 30ms-1.
(b) What will be the flight time of the cannon ball?
Find
time
to
top
of
the
path
Alternative 2 -1
–10ms-1
t t==?? u u= =
30ms
v
=
0
a
=
-1
–
-1
–
30ms
v = 30ms
a = 10ms-2
v = u + vat= u + at
0 = 30– 30
+ =– 10
30 ×
+ t– 10 × t
t t
– 30 =– –60
10= ×– 10
t ×t
6=t
3
=
t
+
So flight
time = 6 seconds
Find total
time
30ms-1
Total time = 2 × 3
= 6 seconds
Vertical Projectile Motion
Example 4
A cannon ball is fired upwards from the ground at a speed
of 30ms-1.
(c) What will be the speed at a height of 30m?
t=?
u = 30ms-1 x = 30m a = –10ms-2
2 = u2 + 2ax
v
v
v2 = 302 + 2 × – 10 × 30
2 = 300
v
v
30m
v =  17.321 ms-1
+
-1 at 30m on
-1
So
the
speed
will
be
17ms
30ms
the way up and 17ms-1 at 30m the way
down
Vertical Projectile Motion
Example 4
A cannon ball is fired upwards from the ground at a speed
of 30ms-1.
(c) What will be the speed at a height of 30m?
Alternative Using
t =energy
? approach
u = 30ms-1 x = 30m a = –10ms-2
-1
-1
v = ? u2 = 30ms
h
=
30m
g
=
10ms
2 + 2ax
v
=
u
v
Eki  Ekf + Ug
v2 =½mu
3022 += ½mv
2 × –2 +
10mgh
× 30
2 = 300
½u2 = ½v2+ gh
v
v
30m
u2 = v2+ 2gh
v = 3017.321
ms-1
2 = v2+ 2×10×30
+
-1 at 30m on
2 will
-1
So
the
speed
be
17ms
900
=
v
+
600
30ms
300and
= v2 17ms-1 at 30m the way
the way up
down 17.3205 = v
v  17 ms-1
Vertical Projectile Motion
Example 5
If a cannon ball fired from 4.0m above the ground is in
the air for 8.0 seconds, what is the launch speed of the
cannon ball?
u = ? x = – 4.0m t = 8.0s
a = –10ms-2
x = ut + ½ at2
– 4 = u × 8 + ½ × – 10 × 82
+
– 4 = 8u – 320
316 = 8u
39.5 = u
4.0m
u  40 ms-1
Vertical Projectile Motion
Exam Questions
2006 Exam
Q6 & 7
When you have F, time and no distance
you
can quite
often
usekg
formulae
A rocket
of mass
0.50
is set onfrom
the ground, pointing vertically up. When
the
momentum/impulse
ignited,
the gunpowderstring
burns for a period of 1.5 s, and provides a constant
forceI of
22 N.
The=mass
= Ft
= p
pf –ofpthe
m(v– u) is very small compared to the mass
i = gunpowder
ofsolve
the rocket,
and can be ignored.
to
the problem
After 1.5 s, what is the height of the rocket above the ground?
𝑭
𝟏𝟕
v=?
t = 1.5s
u=0
a = 𝒏𝒆𝒕 =
= 34ms-2
𝒎
𝟎.𝟓
v = u + at
22N
v = 0 + 34 × 1.5
v = 51 ms-1
+
W = mg
= 0.5× 10
= 5N
alternative using momentum/impulse approach
v=?
t = 1.5s u = 0
Fnet = 17N
Ft = m(v – u)
17 × 1.5 = 0.5v
25.5 = 0.5v
51 ms-1 = v
Horizontally Launched Projectile
Motion
Horizontally Launched Projectile
Motion
Case 1
Case 2
u
Horizontally launched object
drops to ground level
Ex1 – object launched of
building/cliff
Ex2 – object falling off a
moving vehicle
Ex3 – object dropped from
an aircraft
Horizontally Launched Projectile
Motion
Case 1
Case 2
Horizontally launched object
drops to another height
Ex – object launched of
building to the top of
another building
u
u
Case 2: Horizontally Launched
Projectile Motion
Reference
axes
The motion in the x & y
directions are independent
vy is accelerated by g
vx is constant = u
x
In y direction uy = 0 a = 10ms-1
and use const accel equns
In x dirn
y
and use
u
vx
vy
Six points
vx
vy
d
v
If not enough information in
one direction get time from the
other direction
vx
vy
vx = u
If time not involved
Eki + Ug  Ekf
½mu2 + mgh = ½mv2
u2 + 2gh = v2
t
Case 2: Horizontally Launched
Projectile Motion
On landing
thproj=
x
y
u
Rhproj = u
vx
2ℎ
𝑔
v = 𝑢2 + 2𝑔ℎ
vx
vy
 =
vy

Six points
2ℎ
𝑔
vvy
tan-1
2𝑔ℎ
𝑢
What do 2𝑔ℎ
and u represent?
vx & vy
on landing
vx
Four landing formulae
Case 2: Horizontally Launched
Projectile Motion
u
If you treat the top of the second
building as a virtual ground level all
the previous five points including
formulae hold for this situation
h = h 1 – h2
h1
Six points
Virtual ground level
h2
Four landing formulae
Horizontal Projectile Motion
Worked Examples
x
Horizontally Launched Projectile Motion
Example 1
360kmh-1
= 100ms-1
y
An aeroplane travelling at 360kmh-1
drops a food package when it passes
over a hiker at a height of 300m.
(a) How far from the hiker will the
food land?
In x direction
d = ? v = ux = 100ms-1 t = ?7.74597s
d=vt
d = 100 × 7.74597
d = 774.597
d  775m
d=?
Find t from the y direction
t = ? u = 0 x=8.0m a = 10ms-1
x = ut + ½ at2
300= ½ × 10 × t2
300 = 5 × t2
60 = t2
7.74597 = t
x
Horizontally Launched Projectile Motion
Example 1
360kmh-1
= 100ms-1
y
An aeroplane travelling at 360kmh-1
dropped a food package when it passes
over a hiker at a height of 300m.
(a) How far from the hiker will the
food land?
d=?
Alternative
In x direction using derived formulae
Find t from the y direction
-1 t =-1?7.74597s
-1
= ?= ?v = ux = 100ms
t = ? ug==010ms
x=8.0m
a = 10ms-1
Rdhproj
u = 100ms
h = 300m
d=vt
2ℎ
=u
d = 100Rhproj
× 7.74597
𝑔
d = 774.597
2×300
Rhproj = 100
d  775m
10
Rhproj = 774.597
Rhproj  775m
x = ut + ½ at2
300= ½ × 10 × t2
300 = 5 × t2
60 = t2
7.74597 = t
x
Horizontally Launched Projectile Motion
Example 1
An aeroplane travelling at 360kmh-1
dropped a food package when it
passes over a hiker at a height of
300m.
(b) What will be the velocity of the
package when it hits the
ground?
In y direction
vy = ?
uy = 0 x=300m a = 10ms-1
v2 = u2 + 2ax
v2 = 2 × 10 × 300
v2 =6000
v = 77.4597
360kmh-1
= 100ms-1
y
Find final velocity
v=?
100ms-1

v
77.4597ms-1
𝒐𝒑𝒑
𝒂𝒅𝒋
𝟕𝟕.𝟒𝟓𝟗𝟕
tan  =
𝟏𝟎𝟎
𝟕𝟕.𝟒𝟓𝟗𝟕
 = tan-1 𝟏𝟎𝟎
tan  =
h2 = a 2 + b 2
v2 = 1002 + 77.45072
v2 = 16000
v = 126.49
v  126 ms-1
38o
126 ms-1
 = 37.761o
  38o
x
Horizontally Launched Projectile Motion
Example 1
An aeroplane travelling at 360kmh-1
dropped a food package when it
passes over a hiker at a height of
300m.
(b) What will be the velocity of the
package when it hits the
ground?
360kmh-1
= 100ms-1
y
Find final velocity
100ms-1
38o
Alternative using derived formulae
-1
-1 -1
v
=126
10ms
ms
v=?
77.4597ms
38o-1
? u = 100ms
h = 300m g
Inv y=direction
-1
126
ms
-1
vy = ? uy = 0 x=300m a = 10ms
𝒐𝒑𝒑
2𝑔ℎ
2 = a2 + b2-1
2=2
2 + 2ax
h
tan

=

=
tan
v
u
v = 𝑢 + 2𝑔ℎ
𝒂𝒅𝒋
𝑢 2
2 = 1002 + 77.4507
2
v
2 ×210
2×10×4 tan  = 𝟕𝟕.𝟒𝟓𝟗𝟕
-1
v =v2 =100
+×2300
× 10 × 300
2 = 16000
= tan
v
𝟏𝟎𝟎
100
v =6000
v = 126.491
-1 𝟕𝟕.𝟒𝟓𝟗𝟕
o
v = 126.49

=
tan
=
37.761
v = 77.4597
𝟏𝟎𝟎
v  126
mso-1
v  126ms-1
o
 38
 = 37.761
  38o
x
Horizontally Launched Projectile Motion
Example 1
360kmh-1
= 100ms-1
y
An aeroplane travelling at 360kmh-1
dropped a food package when it
passes over a hiker at a height of
300m.
v=?
Find final velocity
(b)
What will be the
velocity
of the for speed (energy
Alternative
if
just
asked
approach)
100ms-1
package when it hits the
ground? v = ? u = 100ms-1 h = 300m g =o 10ms-1
38
Eki + Ug  Ekf
-1
126
ms
2
2
In y direction
½mu + mgh = ½mv v
vy = ? uy = 0 x=300m a = 10ms
½u2 -1+ gh = ½v2
h2 = a2 +2b2
2
v2 = u2 + 2ax
u + 2gh
=v
2 = 1002 + 77.45072
2
v
v = 2 × 10 × 300
2 + 2×10×300 = v2
100
2 = 16000
v
2
v =6000
16000
= v2
v = 126.49
v = 77.4597
v  126= ms
126.491
v -1
v  126ms-1
𝒐𝒑𝒑
𝒂𝒅𝒋
𝟕𝟕.𝟒𝟓𝟗𝟕
tan  =
𝟏𝟎𝟎
𝟕𝟕.𝟒𝟓𝟗𝟕
 = tan-1 𝟏𝟎𝟎
tan  =
 = 37.761o
  38o
x
Horizontally Launched Projectile Motion
Example 1
y
(c) The aeroplane has to do another
drop but from a height 180m. What
is the maximum speed (in kmh-1)
that the plane can have for the food
to land within 300m from the hiker
if the parcel is again released when
the plane is over the hiker?
In x direction
v = ux = ? d= 300m t = ? 6s
v=
𝒅
𝒕
𝟑𝟎𝟎
𝟔
v=
v = 50ms-1
v = 180kmh-1
v = kmh-1
d = 300m
Find t from the y direction
t = ? u = 0 x= 180m a = 10ms-1
x = ut + ½ at2
180= ½ × 10 × t2
180 = 5 × t2
36 = t2
6=t
x
Horizontally Launched Projectile Motion
Example 1
v = kmh-1
y
(c) The aeroplane has to do another
drop but from a height 180m. What
is the maximum speed (in kmh-1)
that the plane can have for the food
to land within 300m from the hiker
if the parcel is again released when
the plane is over the hiker?
d = 300m
Alternative using derived formulae
u In
= ?x direction
Rhproj = 300
v = ux =
R?
v=
-2
h = 180ms-1Findgt =
10ms
from the y direction
2ℎ t = ? 6s
d=
300m
=u
hproj
𝒅
𝒕
𝟑𝟎𝟎
300 =
𝟔
𝑔
2×180
10
𝑢
v=
-1
v = 50ms
300 = 𝑢 × 6
-1
v = 180kmh
50 = u
u = 180kmh-1
t = ? u = 0 x= 180m a = 10ms-1
x = ut + ½ at2
180= ½ × 10 × t2
180 = 5 × t2
36 = t2
6=t
x
Horizontally Launched Projectile Motion
Example 2
20ms-1
h2
y
A car is driven off roof of a 15m
building at 10ms-1 and lands on an
adjacent rooftop 5.0m away? How
high is the adjacent building?
15m
h1
5.0m
Not drawn to scale
Find h2 from x direction
x =? u = 0 a = 10ms-2 t = ?0.5s
x = ut + ½ at2
x = ½ × 10 × 0.52
x = 1.25m
Find height of second building
Height = 15 – 1.25
= 13.75
 14m
Find t from the x direction
t=?
v = ux = 10ms-1 d= 5.0m
t=
𝒅
𝒗
𝟓.𝟎
𝟏𝟎
t=
t = 0.5 s
x
Horizontally Launched Projectile Motion
Example 2
10ms-1
h2
y
A car is driven off roof of a 15m
building at 10ms-1 and lands on an
adjacent rooftop 5.0m away? How
high is the adjacent building?
15m
h1
5.0m
Not drawn to scale
Find h2 from xusing
direction
Alternative
derived formulae
-2 t = ?
-1
-1
x
=?
u
=
0
a
=
10ms
h = h2 = ? Rhproj = 5.0
u0.5s
= 10ms
g
=
10ms
Find t from the x direction
x = ut + ½ at22ℎ
-1
2×ℎ
t
=
?
v
=
u
=
10ms
d= 5.0m
R
=
u
x
2
0.25
=
hproj
x=½
× 10 × 0.5
𝑔
𝒅
10
t
=
x = 1.25m 2×ℎ
1.25 = h𝒗
𝟓.𝟎
5 = 10of second building Find height of second
Find height
building
10
t=
𝟏𝟎
Height = 15 – 1.25
Height
=
15
– s1.25
2×ℎ
t
=
0.5
0.5 = = 13.75
= 13.75
10
 14m
 14m
Horizontal Projectile Motion
Exam Questions
2002 Exam Q5
Q1
A sportscar is driven
horizontally off building
1 and lands on a
building 20m away. The
floor where the car
lands in building 2 is 4.0
m below the floor from
which it started in
building 1.
Calculate the minimum
speed at which the car
should leave building 1
in order to land in the
car park of building .
In x direction
v = ux = ? d= 20m
𝒅
v= 𝒕
𝟐𝟎
v = 𝟎.𝟖𝟗𝟒𝟒𝟑
v = 22.3607ms-1
v  22 ms-1
t = ? 0.89443s
Find t from the y direction
t = ? u = 0 x= 4m a = 10ms-2
x = ut + ½ at2
4= ½ × 10 × t2
4 = 5 × t2
0.8 = t2
0.89443 = t
2002 Exam Q5
Q1
A sportscar is driven
horizontally off building
1 and lands on a
building 20m away. The
floor where the car
lands in building 2 is 4.0
m below the floor from
which it started in
ubuilding
= ? 1.Rhproj = 20.0
Calculate the minimum 2ℎ
= car
u
hproj the
speed at R
which
𝑔
should leave building 1
2×4
in order to land in the
20 = 𝑢
car park of building . 10
Alternative using derived formulae
In x direction
-1
-2
h
=
4.0ms
g
=
10ms
0.89443s
v = ux = ? d= 20m t = ?
𝒅
v= 𝒕
Find t from the y direction
𝟐𝟎
v = 𝟎.𝟖𝟗𝟒𝟒𝟑
t = ? u = 0 x= 4m a = 10ms-2
v = 22.3607ms-1
x = ut + ½ at2
v  22 ms-1
4= ½ × 10 × t2
4 = 5 × t2
0.89443
0.8 = t2
0.89443 = t
20 = 𝑢 ×
22.3607 = u
u  22 ms-1
2002 Exam Q6
Q2
In order to be sure of
landing in the car park of
building 2, Taylor and
Jones in fact left building
1 at a speed of 25 ms–1
Calculate the magnitude
of the velocity of the car
just prior to landing in
the car park of building 2.
In y direction
vy = ? uy = 0 x=4.0m a = 10ms-2
v2 = u2 + 2ax
v2 = 2 × 10 × 4
v2 =80
v = 8.94427
Find final velocity
25ms-1
v
8.94427ms-1
h2 = a2 + b2
v2 = 252 + 8.9944272
v2 = 705
v = 26.5518
v  27 ms-1
2002 Exam Q6
Q2
In order to be sure of
landing in the car park of
building 2, Taylor and
Jones in fact left building
1 at a speed of 25 ms–1
Calculate the magnitude
of the velocity of the car
just prior to landing in
-1 2.
vthe= car
? park
u=
of 25ms
building
Find final velocity
25ms-1

-1
8.94427ms
o
Alternative using Derived Formula
v 38
h = 4.0m
In y direction
2 + 2𝑔ℎ
=
𝑢
-2
vv
=
?
u
y
y = 0 x=4.0m a = 10ms
u22++
2ax
v = v22 =25
2 × 10 × 4
v = 2 × 10 × 4
v =v26.5518
2 =80
-1
v v27ms
= 8.94427
g=
10ms
126 -1ms-1v
h2 = a 2 + b 2
v2 = 252 + 8.9944272
v2 = 705
v = 26.5518
v  27 ms-1
𝒐𝒑𝒑
𝒂𝒅𝒋
𝟖.𝟗𝟒𝟒𝟐𝟕
tan  =
𝟐𝟓
𝒐𝒑𝒑 𝟖.𝟗𝟒𝟒𝟐𝟕
tan = =tan-1
𝒂𝒅𝒋
𝟐𝟓
𝟕𝟕.𝟒𝟓𝟗𝟕 o
tan = =18.96173
o 𝟏𝟎𝟎
  19
𝟕𝟕.𝟒𝟓𝟗𝟕
 = tan-1 𝟏𝟎𝟎
o
tan  =
19 
27 ms-1
= 37.761o
  38o
2002 Exam Q6
Q2
In order to be sure of
landing in the car park of
building 2, Taylor and
Jones in fact left building
1 at a speed of 25 ms–1
Calculate the magnitude
of the velocity of the car
just prior to landing in
-1 2.
vthe= car
? park
u=
of 25ms
building
Find final velocity
25ms-1

-1
8.94427ms
o
Alternative using energy approach
v 38
h = 4.0m
In y direction
i+ U  E f
E
k
g
ka = 10ms-2
vy = ? uy = 0 x=4.0m
2 2 mgh = ½mv2
½mu
v2 = +
u + 2ax
v2 =u22 ×+102gh
× 4 = v2
v2 =80 2
𝑢 + 2𝑔ℎ
vv==8.94427
𝒐𝒑𝒑
𝒂𝒅𝒋
𝟖.𝟗𝟒𝟒𝟐𝟕
h2 = a 2 + b 2
tan  =
𝟐𝟓
v2 = 252 + 8.99442722 tan  = 𝒐𝒑𝒑
-1 𝟖.𝟗𝟒𝟒𝟐𝟕

=
tan
𝒂𝒅𝒋
𝟐𝟓
v2 = 705
𝟕𝟕.𝟒𝟓𝟗𝟕 o
v = 26.5518
tan = =18.96173
o 𝟏𝟎𝟎
-1
  19
𝟕𝟕.𝟒𝟓𝟗𝟕
v  27 ms-1
 = tan-1 𝟏𝟎𝟎
o
g=
-1 -1
v
10Nkg
126 ms
tan  =
v = 25 + 2 × 10 × 4
v = 26.5518
v  27ms
19  = 37.761o
Same as derived
formula
-1
  38o
27 ms
2006 Exam Q8
Q3
A 0.5kg rocket is launched horizontally and propelled by a constant force of
22 N for 1.5 s, from the top of a 50 m tall building. Assume that in its
subsequent motion the rocket always points horizontally.
After 1.5 s, what is the speed of the rocket, and at what angle is the rocket
moving relative to the ground?
In x direction
𝑭
𝟐𝟐
22N
v=?
t = 1.5s
u=0
a = 𝒏𝒆𝒕 =
= 44ms-2
𝒎
𝟎.𝟓
v = u + at
v = 0 + 44 × 1.5
v = 66 ms-1
x
alternative
using momentum/impulse approach
In
y direction
1.5s uu == 00
Fanet
= 22N-2
v=?
t t==1.5s
= 10ms
y
vFt==um(v
+ at– u)
v22= ×0 1.5
+ 10= 0.5v
× 1.5
v33= =150.5v
ms-1
66 ms-1 = v
2006 Exam Q8
Q3
After 1.5 s, what is the speed of the rocket, and at what angle is the rocket
moving relative to the ground?
Find final velocity
66ms-1

22N
v
x
y
15ms-1
𝒐𝒑𝒑
𝒂𝒅𝒋
𝟏𝟓
tan  =
𝟔𝟔
𝟏𝟓
 = tan-1 𝟔𝟔
tan  =
h2 = a 2 + b 2
v2 = 662 + 152
v2 = 4581
v = 67.6831
v  68 ms-1
 = 12.8043o
  13o
13o
68 ms-1
Obliquely Launched Projectile
Motion
Obliquely Launched Projectile Motion
Case 1: landing height = launch height
Case 2: landing height different to launch height
Case 1: Obliquely Launched Projectiles Landing At The
Launching Height
The motion in the x & y
directions are independent
uy = u sin 
vx is constant = u cos 
In y direction
uy = u sin a = –10ms-1
use const accel equns
In x dirn
vx = u cos in
If time not involved
Eki  Ekf + Ug
½mu2 = ½mv2+ mgh
vxu2 = v2+ 2gh
vx
y

vy
u
vx
Ekmin = ½ mvx2
uy = u sin 
ux = u cos 
t
v
If not enough info in one dirn
get time from the other dirn
Reference
axes
x
d
Speeds before hmax are
theEight
same as points
those after
at each height
vy
vx
=
v = vuy
Case 1: Obliquely Launched Projectiles Landing At The
Launching Height
𝑢2 𝑠𝑖𝑛2 𝜃
hmax=
2𝑔
𝑢2 𝑠𝑖𝑛2𝜃
range =
𝑔
2𝑢𝑠𝑖𝑛𝜃
tflight =
𝑔
vx
y
x

vy
u
vx
uy = u sin 
ux = u cos 
vx
vy
vx
=
Three formulae
v=u
vy
Case 2: Obliquely Launched Projectiles Landing At A
Different Height To The Launching Height
Notes for Case 2
• The eight points relating to oblique projectiles that land at the
launching height also apply to this type of situation.
• You will not be asked to find the time from the vertical axis
since it will involve solving a quadratic equation. Instead you
will either be given the time or asked to work it out from a
𝑑
horizontal distance (using t = )
𝑣
u
x
Obliquely Launched Projectile
Motion
Worked Examples
Obliquely Launched Projectile Motion
Example 1
y
What are the three other ways of working out the
flight time?
1. t = ? u = 13.101x x31ms
=0 a-1= –10ms-2 (this will
involve
A golf ball is hit from the ground
ata factorisation)
2. t = ? u = 13.101 v = u–13.101
a = –10ms-2
31ms-1 at 25o to the horizontal.
y = 31sin25
-1
(this is pretty easy)
=
13.101ms
o
25
(a) How far will the ball travel
3. Using flight time derived formula (just put the
before it strikes the groundnumbers
if
uinx and
= 31cos25
d
work out)
the ground is flat?
In x direction
d = ? v = 28.096ms-1 t = ?
= 28.096ms-1
Find t to top of path from y dirn
t = ? u = 13.101 v =0 a = –10ms-2
v = u + at
0 = 13.101 + – 10 × t
– 13.101 = – 10 × t
1.3101 = t
Find total time
Total time = 2 × 1.3101
= 2.6202s
Obliquely Launched Projectile Motion
Example 1
y
What are the three other ways of working out the
flight time?
uy-2=(this
31sin25
1. t = ? u = 13.101x x =0 a = –10ms
will
= 13.101ms-1
involve
A golf ball is hit from the ground
ata factorisation)
2. t = ? u = 13.101 v = –13.101 a = –10ms-2
31ms-1 at 25o to the horizontal.
(this is pretty easy)
ux = 31cos25
(a) How far will the ball travel
3. Using flight time derived
formula (just put the
before it strikes the groundnumbers
if
28.096ms-1
in and work=out)
the ground is flat?
Find t from y dirn
In x direction
d = ? v = 28.096ms-1 t = ?
t = ? u = 13.101 x =-0
a = –10ms-2
x = ut + ½at2
0 = 13.101 t + ½ × – 10 × t2
0 = 13.101 t – – 5 t2
0 = t (13.101 – 5t )
𝟏𝟑.𝟏𝟎𝟏
t = 0,
𝟓
t = 2.6202
Obliquely Launched Projectile Motion
Example 1
y
What are the three other ways of working out the
flight time?
uy-2=(this
31sin25
1. t = ? u = 13.101x x =0 a = –10ms
will
= 13.101ms-1
involve
A golf ball is hit from the ground
ata factorisation)
2. t = ? u = 13.101 v = –13.101 a = –10ms-2
31ms-1 at 25o to the horizontal.
(this is pretty easy)
ux = 31cos25
(a) How far will the ball travel
3. Using flight time derived
formula (just put the
before it strikes the groundnumbers
if
28.096ms-1
in and work=out)
the ground is flat?
Find t from y dirn
In x direction
d = ? v = 28.096ms-1 t = ?
t = ? u = 13.101 v =-13.101 a = –10ms-2
v = u + at
-13.101 = 13.101 + – 10 × t
– 26.202 = – 10 × t
2.6202 = t
Obliquely Launched Projectile Motion
Example 1
y
What are the three other ways of working out the
flight time?
uy-2=(this
31sin25
1. t = ? u = 13.101x x =0 a = –10ms
will
= 13.101ms-1
involve
A golf ball is hit from the ground
ata factorisation)
2. t = ? u = 13.101 v = –13.101 a = –10ms-2
31ms-1 at 25o to the horizontal.
(this is pretty easy)
ux = 31cos25
(a) How far will the ball travel
3. Using flight time derived
formula (just put the
before it strikes the groundnumbers
if
28.096ms-1
in and work=out)
the ground is flat?
In x direction
d = ? v = 28.096ms-1 t = ?2.6202s
d=vt
d = 28.096 × 2.6202
d = 73.617
d  74m
Find t from flight formula
t = ? u = 31ms-1  = 25o g = –10ms-2
2𝑢𝑠𝑖𝑛𝜃
tflight =
𝑔
2×31×𝑠𝑖𝑛25
tflight =
10
tflight = 2.6202
Obliquely Launched Projectile Motion
Example 1
A golf ball is hit from the ground at
31ms-1 at 25o to the horizontal.
(a) How far will the ball travel
before it strikes the ground if
the ground is flat?
31ms-1
25o
range
Find t to top
of path from y dirn
1 step alternative using derived
formulae
In x direction
? u =-113.101 v =0 a = –10ms-2
Robliq = ? u = 31ms-1-1  =2.6202s
25o gt == 10ms
d = ? v = 28.096ms t = ?
v = u + at
𝑢2 𝑠𝑖𝑛2𝜃
d=vt
0 = 13.101 + – 10 ×
Robliq =
𝑔
– 13.101 = – 10 × t
d = 28.096 × 2.6202
2
31 sin(2×25)
R
=
d = 73.617 obliq
1.3101 = t
10
Find total time
d  74m Robliq = 73.617
Robliq  74m
Total time = 2 × 1.3101
= 2.6202s
Obliquely Launched Projectile Motion
Example 1
y
x
31ms-1
A golf ball is hit from the ground at
uy = 31sin25
31ms-1 at 25o to the horizontal.
-1
=
13.101ms
o
25
(b) What is the minimum speed of
the golf ball?
ux = 31cos25
= 28.096ms-1
Minimum speed occurs at the top
of the path when there only the
horizontal component of velocity.
So minimum is 28.096ms-1
Obliquely Launched Projectile Motion
Example 1
y
A golf ball is hit from the ground
at 31ms-1 at 25o to the horizontal.
(c) How high will the golf ball
rise?
x
31ms-1
uy = 31sin25
= 13.101ms-1
25o
ux = 31cos25
= 28.096ms-1
In y direction
x=?
u = 13.101ms-1 v = 0
a = –10ms-2
v2 = u2 + 2ax
0 = 13.1012 + 2 × – 10 × x
0 = 171.636 + – 20 × x
– 171.636 = – 20 × x
8.5818 = x
x = 8.6m
Obliquely Launched Projectile Motion
Example 1
y
A golf ball is hit from the ground
at 31ms-1 at 25o to the horizontal.
(c) How high will the golf ball
rise?
x
31ms-1
uy = 31sin25
= 13.101ms-1
25o
ux = 31cos25
= 28.096ms-1
Alternative
In y direction using derived formulae
-1
-2
31ms-1 -1
 = 25
u = 13.101ms
v o= 0g = 10ms
a = –10ms
𝑢22𝑠𝑖𝑛2 𝜃
2
hvmax=
= u + 2ax
2𝑔
2
–
0 = 13.101
312 𝑠𝑖𝑛2 25 + 2 × 10 × x
h =
2×10
0max
= 171.636
+ – 20 × x
hmax==8.5820
– 171.636
– 20 × x
hmax=x8.6m
8.5818
x = 8.6m
hxmax
= =? ?
Obliquely Launched Projectile Motion
Example 1
y
x
31ms-1
A golf ball is hit from the ground at
uy = 31sin25
31ms-1 at 25o to the horizontal.
-1
=
13.101ms
o
25
(d) What is the velocity of the ball
as it hits the ground?
ux = 31cos25
= 28.096ms-1
Since the ball hits the ground at
the same height as it launch
height final speed is:
25o
31ms-1
Obliquely Launched Projectile Motion
Example 1
y
x
A golf ball is hit from the ground at
31ms-1 at 25o to the horizontal.
(e) What will be the speed of the
ball 20m off the ground?
31ms-1
25o
Using energy approach (which avoids having to work with vectors)
v=?
u = 31ms-1
h = 20m
Eki  Ekf + Ug
½mu2 = ½mv2 + mgh
½u2 = ½v2+ gh
u2 = v2+ 2gh
312 = v2+ 2×10×20
961 = v2 + 400
561 = v2
23.685 = v
v  24 ms-1
g = 10ms-1
Obliquely Launched Projectile Motion
Example 2
y
x
A golf ball is hit at 25ms-1 at 35o to the
horizontal from a cliff edge and it takes
6.0 seconds to hit the ground below the
cliff.
(a) How high is the cliff?
x
In y direction
25ms-1
x=?
u = 14.3390ms-1 t = 6.0s
a = –10ms-2
x = ut + ½ at2
x = 14.3390 × 6 + ½ × – 10 × 62
x = – 93.966
so the cliff is  94m high
uy = 25sin35
= 14.339ms-1
35o
ux = 25cos35
= 20.4796ms-1
Obliquely Launched Projectile Motion
Example 2
y
A golf ball is hit at 25ms-1 at 35o to the
horizontal from a cliff edge and it takes
6.0 seconds to hit the ground below the
cliff.
(b) How far from the base of the cliff
does the ball land?
In x direction
d = ? v = 20.4796ms-1 t = 6
d=vt
d = 20.4796 × 6
d = 122.875
d  123m
x
d
25ms-1
uy = 25sin35
= 14.339ms-1
35o
ux = 25cos35
= 20.4796ms-1
y
Obliquely Launched Projectile Motion
Example 3
x
28ms-1
60o
A golf ball is hit at
at
to the
horizontal from a fairway and hits a tree that is
56m away. How far up the tree does the golf
ball strike the tree?
-1
28ms
28ms-1
h
60o
56m
uy = 28sin60
= 24.249ms-1
60o
ux = 28cos60
= 14ms-1
In y direction
x = ? u = 24.249ms-1 a = –10ms-2 t = ?4s
x = ut + ½ at2
x = 24.249 × 4 + ½ × – 10 × 42
x = 16.996
so the ball hits the tree  17m up
Find t from the x direction
t = ? v = ux = 14ms-1 d= 56m
𝒅
t=
𝒗
𝟓𝟔
𝟏𝟒
t=
t=4s
Obliquely Launched Projectile
Motion
Exam Questions
2004 Exam Q7
Q1
The diagram shows a motorcycle rider using a 20° ramp to jump her
motorcycle across a river that is 10.0 m wide.
Question 7
Calculate the minimum speed that the motorcycle and rider must leave the
top of the first ramp to cross safely to the second ramp that is at the same
height. (The motorcycle and rider can be treated as a point-particle.)
u=?
Robliq =10
 = 20o g = 10ms-2
𝑢2 𝑠𝑖𝑛2𝜃
Robliq =
𝑔
Using x & y direction analysis for this
𝑢 sin(2×20)
problem
is very difficult so using the
10 =
10
range
is a much more practical
100 =formula
u2sin40
155.572
= u2
way
to solve
this problem
2
12.4729 = u
u  12 ms-1
30ms-1
2005 Exam Q11
y
Q2
3.0m
x
8o
30ms-1
A ball leaves a racket 3.0 m above the ground at at an angle of 8° to the
horizontal. At its maximum height it has a speed of 30.0 m s-1.
With what speed, relative to the deck, did the ball leave Fred's racket? Give
your answer to three significant figures.
cos  =
u
8o
30ms-1
cos 8 =
u=
𝑎𝑑𝑗
ℎ𝑦𝑝
30
𝑢
30
𝑐𝑜𝑠8
u = 30.29483ms-1
u  30.3ms-1
30ms-1
2005 Exam Q12
y
Q3
3.0m
x
x
8o
30ms-1
3.0m
A ball leaves a racket 3.0 m above the ground at at an angle of 8° to the
horizontal. At its maximum height it has a speed of 30.0 m s-1.
At its highest point, how far was the ball above the ground.
x=?
30.29483ms-1
8o
30ms-1
u = 30ms-1 v = 0 a = –10ms-2
v2 = u2 + 2ax
0 = 4.2162252 + 2 × – 10 × x
0 = 17.77655 + – 20 × x
– 17.77655 = – 20 × x
0.888828 = x
30.29483sin8
= 4.216225ms-1
Overall height = 0.888828 + 3
= 3.888828
 3.89 m
30ms-1
2005 Exam Q12
y
Q3
3.0m
x
hmax
8o
30ms-1
3.0m
A ball leaves a racket 3.0 m above the ground at at an angle of 8° to the
horizontal. At its maximum height it has a speed of 30.0 m s-1.
At its highest point, how far was the ball above the ground.
Alternative
derived
-1
x = ?using
u = 30ms
v formulae
= 0 a = –10ms-2
-1
hmax= ? u = 30.29483ms
 = 8o g = 10ms-2
v2 = u2 + 2ax
2
30.29483ms-1
8o
30ms-1
2
𝑢 𝑠𝑖𝑛 𝜃
hmax=0 = 4.2162252 + 2 × – 10 × x
2𝑔
–
020.29483
= 17.77655
2 𝑠𝑖𝑛2+
8 20 × x
hmax=– 17.77655 = – 20 × x
2×10
0.888828
=x
hmax =
0.888828
Overall height = 0.888828 + 3
= 3.888828
 3.89 m
30ms-1
2005 Exam Q12
y
x
Q3
h1 = 3.0m
8o
30ms-1
h2
A ball leaves a racket 3.0 m above the ground at at an angle of 8° to the
horizontal. At its maximum height it has a speed of 30.0 m s-1.
At its highest point, how far was the ball above the ground.
Alternative using Energy Approach
30.29483ms-1
8o
30ms-1
-1
–
-2
-1
= ? u = 30ms
= 0 ga == 10ms
10ms-2
u =x 30.29483ms
h1 =v 3m
v2 =f u2 + 2ax
i
i
Ek + Ug  Ek + Ugf 2
0 = 4.216225 + 2 × – 10 × x
2 + mgh
½ mu2 + mgh1 = 0½=mv
17.77655 +2 – 20 × x
½ u2 + gh1 = –½17.77655
v2 + gh2= – 20 × x
½ × 30.294832 + 100.888828
× 3 = ½ =× x302 + 10 × h2
488.888 = 450 + 10 × h
Overall height = 0.8888282 + 3
3.8888 = h2 = 3.888828
h2= ?
h2  3.89m
 3.89 m
2007 Exam Q14
Q4
Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of
40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the
ball hits John and the height from which the ball was fired are the same.
What is the time of flight of the paintball?
40ms-1
y
uy = 40sin25
= 16.9047ms-1
x
25o
ux = 40cos25
= 36.2523ms-1
Find t from the x direction
t = ? v = ux = 28.096ms-1 d= 127m
𝒅
t=
𝒗
𝟏𝟐𝟕
t=
𝟐𝟖.𝟎𝟗𝟔
t = 3.5032 s
t  3.5 s
2007 Exam Q14
Q4
Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of
40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the
ball hits John and the height from which the ball was fired are the same.
What is the time of flight of the paintball?
Alternative
40ms-1
y
x
Find t from the y direction
t = ? u = 16.9047ms-1 v = –16.9047ms-1 a = –10ms-2
uy = 40sin25
= 16.9047ms-1
25o
ux = 40cos25
= 36.2523ms-1
v = u + at
- 16.9047 = 16.9047 + – 10 × t
– 33.8094 = – 10 × t
3.38094 = t
t  3.4 s
There is a discrepancy in the data for this
problem hence the slight difference in
answers from the x direction and y direction
2007 Exam Q14
Q4
Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of
40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the
ball hits John and the height from which the ball was fired are the same.
What is the time of flight of the paintball?
40ms-1
y
x
Find t from the y direction
t = ? u = 16.9047ms-1 v = –16.9047ms-1 a = –10ms-2
uy = 40sin25
= 16.9047ms-1
25o
ux = 40cos25
= 36.2523ms-1
v = u + at
- 16.9047 = 16.9047 + – 10 × t
– 33.8094 = – 10 × t
3.38094 = t
t  3.4 s
There is a discrepancy in the data for this
problem hence the slight difference in
answers from the x direction and y direction
2007 Exam Q14
Q4
Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of
40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the
ball hits John and the height from which the ball was fired are the same.
What is the time of flight of the paintball?
Alternative with formula
40ms-1
y
x
Find t from flight formula
uy = 40sin25
= 16.9047ms-1
25o
ux = 40cos25
= 36.2523ms-1
t = ? u = 40ms-1  = 25o g = –10ms-2
2𝑢𝑠𝑖𝑛𝜃
𝑔
2×40×𝑠𝑖𝑛25
tflight =
10
tflight =
tflight = 3.3809
tflight  3.4s
There is a discrepancy in the data for this
problem hence the slight difference in
answers from the previous methods
2007 Exam Q15
Q5
Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of
40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the
ball hits John and the height from which the ball was fired are the same.
What is the value of h, the maximum height above the firing level?
40ms-1
y
Find hmax from y direction
u = 16.9047ms-1 v = 0 a = –10ms-2
v2 = u2 + 2ax
0 = 16.90472 + 2 × – 10 × x
0 = 287.201+ – 20 × x
– 287.201 = – 20 × x
14.3600 = x
x  14m
x=?
x
uy = 40sin25
= 16.9047ms-1
25o
ux = 40cos25
= 36.2523ms-1
2007 Exam Q15
Q5
Daniel fires a ‘paintball’ at an angle of 25° to the horizontal and a speed of
40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the
ball hits John and the height from which the ball was fired are the same.
What is the value of h, the maximum height above the firing level?
Alternative with formula
40ms-1
y
Find t from flight formula
uy = 40sin25
= 16.9047ms-1
x
25o
ux = 40cos25
= 36.2523ms-1
hmax= ?
u = 40ms-1
 = 25o g = 10ms-2
𝑢2 𝑠𝑖𝑛2 𝜃
hmax=
2𝑔
402 𝑠𝑖𝑛2 25
hmax=
2×10
hmax = 14.2885
hmax  14m
There is a discrepancy in the data for this
problem hence the slight difference in
answers from the previous method
2007 Exam Q16
Q6
Daniel fires a ‘paintball’ at an angle of
25° to the horizontal and a speed of 40.0
ms–1. The paintball hits John, who is 127
m away. The height at which the ball hits
John and the height from which the ball
was fired are the same.
Which of the following diagrams (A–D)
below gives the direction of the force
acting on the paintball at points X and Y
respectively?
Since air resistance is ignored the
only force on the paintball is gravity
which is constant and downwards,
so the acceleration must be
constant and downwards.
2007 Exam Q17
Q7
Later in the game, Daniel is twice as far away from John (254 m). John fires an
identical paintball from the same height above the ground as before. The ball
hits Daniel at the same height as before. In both cases the paintball reaches the
same maximum height (h) above the ground.
Which one or more of the following is the same in both cases?
A.
flight time
Because the paintball gets to the same
vertical height the vertical component in each
B.
initial speed
situation must be the same and hence the
C.
acceleration
flight time will also be the same.
D.
angle of firing level?
Since air resistance is ignored the
only force on the paintball is gravity
which is constant in both situations
and so the acceleration is constant
as well.
So With More Than One Method
For Solving Most Projectile
Motion Questions – How Do I
Decide On The Best?
Some Loose Rules To Follow.
Some Loose Rules For Deciding On
The Best Method To Solve A Problem
1. The best method is the one that makes the most sense
to you and doesn’t take forever to get to an answer
2. Often if there is a formula use a formula, but be
careful of expressions such as
•
20.294832 𝑠𝑖𝑛2 8
hmax=
= 0.888828
2×10
𝑢2 𝑠𝑖𝑛2𝜃
Robliq = 𝑔
v = 1002 + 2 × 10 × 300 = 126.491
•
Rhproj = 100
•
•
2×300
10
= 774.597
3. Make sure you practice more than one method when
solving projectile motion questions – particularly
working with x & y directions.
Some Loose Rules For Deciding On
The Best Method To Solve A Problem
4. An energy approach is very useful when asked for the
magnitude of a velocity that is not at launching height
or maximum height.
5. Remember there are three approaches that can be used
for solving
Constant Accel Equations
Energy
Formulae
(no time)
Impulse/
Momentum
Formulae
(no dist)