1. Projectile Motion Notes

2. Vertical Projectile Motion

3. Vertical Projectile Motion
Case Case Case Case 4 u
Object launched vertically and returns to launch height
Object launched vertically and returns to a different height
Object dropped from rest
Object projected downwards u u

4. 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

5. Case 2: Vertical Projectile Motion+
Set positive direction as down u
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

6. 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
Speeds up are the same as down at each height
Use const accel equations with a = -10ms-1 v v +
Time up = time down or
Total time = 2 × time up
Set positive direction as up u
Six points

7. Case 4: Vertical Projectile Motion
If time not involved can use energy approach
Eki  Ekf + Ug
½mu2 = ½mv2 + mgh
v = 0
Speeds up are the same as down at each height
Use const accel equations with a = -10ms-1 u +
Displacements below launch height will be negative
Set positive direction as up
Six points

8. Case 4: Vertical Projectile Motion
If time not involved can use energy approach
Eki  Ekf + Ug
½mu2 = ½mv2 + mgh
v = 0 u
Speeds up are the same as down at each height
Use const accel equations with a = -10ms-1 +
Displacements below launch height will be negative
Set positive direction as up
Six points

9. Vertical Projectile Motion
Worked Examples

10. 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 8 = ½ × 10 × t2 8 = 5 × t2 1.6 = t = t t  1.3 s
In Maths when you determine the square root of a number you should always give the result as a  value. This means there are two answers
In Physics we normally just write out the magnitude of the square root because the negative value is either not relevant or it represents a direction which is already obvious in the problem. In this case a negative time is not plausible. +

11. 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 v2 = 2 × 10 × 8 v2 = 160 v = v  13 ms-1
Alternative – Energy Approach
Ug  Ek
mgh = ½mv2
gh = ½v2
2gh = v2
2 × 10 × 8 = v2
160 = v2
= v
v = 13ms-1 +
In In Physics we normally just write out the magnitude of the square root because the negative value is either not relevant or it represents a direction which is already obvious in the problem. In this case a negative value means the stone is moving upwards which is not plausible.

12. 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 a = 10ms-2 x = ut + ½ at2 x = ½ × 10 × 2.52 x = x  31 s +

13. 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 v2 = u2 + 2ax v2 = × 10 × 3 v2 = 64 v = 8 ms-1 +
Alternative – Energy Approach
Eki + Ug  Ekf
½mu2 + mgh = ½mv2
½u2 + gh = ½v2
u2 + 2gh = v2
× 10 × 3 = v2
64 = v2
8 = v
v = 8ms-1
2.0ms-1

14. 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 = u2 + 2ax 0 = × – 10 × x 0 = – 20 × x – 900 = – 20 × x 45 = x x = 45m
30ms-1 x +

15. 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 = u2 + 2ax 0 = × – 10 × x 0 = – 20 × x – 900 = – 20 × x 45 = x x = 45m
30ms-1
Alternative – Energy Approach
EkUg
½mv2 = mgh
½v2 = gh
v2 = 2gh
302 = 2 × 10 × h
900 = 20h
45 = h
h = 45m x +

16. 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 – 30 = – 10 × t 3 = t Find total time Total time = 2 × 3 = 6 seconds
30ms-1 t t +

17. 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-1 v = u + at 0 = 30 + – 10 × t – 30 = – 10 × t 3 = t Find total time Total time = 2 × 3 = 6 seconds
Alternative 1
t = ? u = 30ms x = 0 a = – 10ms-2
x = ut + ½at2
0 = 30t + ½ × – 10 × t2
0 = 30t + – 5 × t2
0 = 5t (6 – t)
t = 0, 6
So flight time = 6 seconds
30ms-1 t t +

18. 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-1 v = u + at 0 = 30 + – 10 × t – 30 = – 10 × t 3 = t Find total time Total time = 2 × 3 = 6 seconds
Alternative 2
t = ? u = 30ms v = – 30ms-1 a = – 10ms-2
v = u + at
– 30 = – 10 × t
– 60 = – 10 × t
6 = t
So flight time = 6 seconds
30ms-1 t t +

19. 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 v2 = u2 + 2ax v2 = × – 10 × 30 v2 = 300 v =  ms-1 So the speed will be 17ms-1 at 30m on the way up and 17ms-1 at 30m the way down
30ms-1 v v
30m +

20. 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 v2 = u2 + 2ax v2 = × – 10 × 30 v2 = 300 v =  ms-1 So the speed will be 17ms-1 at 30m on the way up and 17ms-1 at 30m the way down
Alternative Using energy approach
v = ? u = 30ms h = 30m g = 10ms-1
Eki  Ekf + Ug
½mu2 = ½mv2 + mgh
½u2 = ½v2+ gh
u2 = v2+ 2gh
302 = v2+ 2×10×30
900 = v
300 = v2
= v
v  17 ms-1
30ms-1 v v
30m +

21. 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 – = 8u 39.5 = u u  40 ms-1 +
4.0m

22. Vertical Projectile Motion
Exam Questions

23. + I = Ft = p = pf – pi = m(v– u) 22N W = mg = 0.5× 10 = 5N
2006 Exam Q6 & 7
When you have F, time and no distance you can quite often use formulae from the momentum/impulse string
I = Ft = p = pf – pi = m(v– u)
to solve the problem
A rocket of mass 0.50 kg is set on the ground, pointing vertically up. When ignited, the gunpowder burns for a period of 1.5 s, and provides a constant force of 22 N. The mass of the gunpowder is very small compared to the mass of the rocket, and can be ignored. After 1.5 s, what is the height of the rocket above the ground?
v = ? t = 1.5s u = 0 a = 𝑭 𝒏𝒆𝒕 𝒎 = 𝟏𝟕 𝟎.𝟓 = 34ms-2
v = u + at
v = × 1.5
v = 51 ms-1
alternative using momentum/impulse approach
v = ? t = 1.5s u = Fnet = 17N
Ft = m(v – u)
17 × 1.5 = 0.5v
25.5 = 0.5v
51 ms-1 = v
22N +
W = mg
= 0.5× 10
= 5N

24. Horizontally Launched Projectile Motion

25. Horizontally Launched Projectile Motion
Case Case 2
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 u u u

26. Horizontally Launched Projectile Motion
Case Case 2
Horizontally launched object drops to another height
Ex – object launched of building to the top of another building u u

27. 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
In y direction uy = 0 a = 10ms-1
and use const accel equns y x
In x dirn vx = u
and use d v t u vx vy
If not enough information in one direction get time from the other direction vx vy
If time not involved
Eki + Ug  Ekf
½mu2 + mgh = ½mv2
u2 + 2gh = v2 vx vy
Six points

28. Case 2: Horizontally Launched Projectile Motion
On landing
thproj= 2ℎ 𝑔 y x
Rhproj = u 2ℎ 𝑔 u vx vy
v = 𝑢 2 +2𝑔ℎ
What do 2𝑔ℎ and u represent?
vx & vy
on landing vx vy
 = tan 𝑔ℎ 𝑢 vx  vy
Six points
Four landing formulae v

29. Case 2: Horizontally Launched Projectile Motion
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 u
h = h1 – h2 h1
Virtual ground level h2
Six points
Four landing formulae

30. Horizontal Projectile Motion
Worked Examples

31. Horizontally Launched Projectile Motion Example 1
360kmh-1
= 100ms-1
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 = ? d = v t d = 100 × d = 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
= t s

32. Horizontally Launched Projectile Motion Example 1
360kmh-1
= 100ms-1
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? In x direction d = ? v = ux = 100ms-1 t = ? d = v t d = 100 × d = d  775m
d = ?
Alternative using derived formulae
Rhproj = ? u = 100ms h = 300m g = 10ms-1
Rhproj = u 2ℎ 𝑔
Rhproj = ×300 10
Rhproj =
Rhproj  775m
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
= t s

33. Horizontally Launched Projectile Motion Example 1
360kmh-1
= 100ms-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 =
Find final velocity
h2 = a2 + b2
v2 =
v2 = 16000
v =
v  126 ms-1
v = ?
100ms-1 
ms-1 v
tan  = 𝒐𝒑𝒑 𝒂𝒅𝒋
tan  = 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = tan-1 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = o
  38o
38o
126 ms-1

34. Horizontally Launched Projectile Motion Example 1
360kmh-1
= 100ms-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 =
Find final velocity
h2 = a2 + b2
v2 =
v2 = 16000
v =
v  126 ms-1
v = ?
100ms-1
Alternative using derived formulae
v = ? u = 100ms h = 300m g = 10ms-1
v = 𝑢 2 +2𝑔ℎ
v = ×10×300
v =
v  126ms-1 
38o
ms-1
38o
126 ms-1 v
126 ms-1
 = tan 𝑔ℎ 𝑢
 = tan ×10×4 100
 = o
  38o
tan  = 𝒐𝒑𝒑 𝒂𝒅𝒋
tan  = 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = tan-1 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = o
  38o

35. Horizontally Launched Projectile Motion Example 1
360kmh-1
= 100ms-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 =
v = ?
Find final velocity
h2 = a2 + b2
v2 =
v2 = 16000
v =
v  126 ms-1
Alternative if just asked for speed (energy approach)
v = ? u = 100ms h = 300m g = 10ms-1
Eki + Ug  Ekf
½mu2 + mgh = ½mv2
½u2 + gh = ½v2
u2 + 2gh = v2
×10×300 = v2
16000 = v2
= v
v  126ms-1
100ms-1 
38o
126 ms-1 v
tan  = 𝒐𝒑𝒑 𝒂𝒅𝒋
tan  = 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = tan-1 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = o
  38o

36. Horizontally Launched Projectile Motion Example 1
v = kmh-1
(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 = ? v = 𝒅 𝒕 v = 𝟑𝟎𝟎 𝟔 v = 50ms-1 v = 180kmh-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 6s

37. Horizontally Launched Projectile Motion Example 1
v = kmh-1
(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 = ? v = 𝒅 𝒕 v = 𝟑𝟎𝟎 𝟔 v = 50ms-1 v = 180kmh-1
d = 300m
Alternative using derived formulae
u = ? Rhproj = h = 180ms g = 10ms-2
Rhproj = u 2ℎ 𝑔
300 = 𝑢 2×180 10
300 = 𝑢 × 6
50 = u
u = 180kmh-1
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 6s

38. Horizontally Launched Projectile Motion Example 2
20ms-1
15m h2
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?
Find h2 from x direction
x =? u = 0 a = 10ms-2 t = ?
x = ut + ½ at2
x = ½ × 10 × 0.52
x = 1.25m
Find height of second building
Height = 15 – 1.25
= 13.75
 14m h1
5.0m
Not drawn to scale
0.5s
Find t from the x direction
t = ? v = ux = 10ms d= 5.0m
t = 𝒅 𝒗
t = 𝟓.𝟎 𝟏𝟎
t = 0.5 s

39. Horizontally Launched Projectile Motion Example 2
10ms-1
15m h2
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?
Find h2 from x direction
x =? u = 0 a = 10ms-2 t = ?
x = ut + ½ at2
x = ½ × 10 × 0.52
x = 1.25m
Find height of second building
Height = 15 – 1.25
= 13.75
 14m h1
5.0m
Not drawn to scale
Alternative using derived formulae
h = h2 = ? Rhproj = u = 10ms g = 10ms-1
Rhproj = u 2ℎ 𝑔
5 = ×ℎ 10
0.5 = 2×ℎ 10
0.5s
Find t from the x direction
t = ? v = ux = 10ms d= 5.0m
t = 𝒅 𝒗
t = 𝟓.𝟎 𝟏𝟎
t = 0.5 s
0.25 = 2×ℎ 10
1.25 = h
Find height of second building
Height = 15 – 1.25
= 13.75
 14m

40. Horizontal Projectile Motion
Exam Questions

41. 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 t = ?
v = 𝒅 𝒕
v = 𝟐𝟎 𝟎.𝟖𝟗𝟒𝟒𝟑
v = ms-1
v  22 ms-1 s
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
= t

42. Alternative using derived formulae
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 .
Alternative using derived formulae
u = ? Rhproj = h = 4.0ms g = 10ms-2
Rhproj = u 2ℎ 𝑔
20 = 𝑢 2×4 10
20 = 𝑢 ×
= u
u  22 ms-1
In x direction
v = ux = ? d= 20m t = ?
v = 𝒅 𝒕
v = 𝟐𝟎 𝟎.𝟖𝟗𝟒𝟒𝟑
v = ms-1
v  22 ms-1 s
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
= t

43. 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.
Find final velocity
h2 = a2 + b2
v2 =
v2 = 705
v =
v  27 ms-1
25ms-1
ms-1 v
In y direction
vy = ? uy = 0 x=4.0m a = 10ms-2
v2 = u2 + 2ax
v2 = 2 × 10 × 4
v2 =80
v =

44. Alternative using Derived Formula
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.
Find final velocity
h2 = a2 + b2
v2 =
v2 = 705
v =
v  27 ms-1
25ms-1 
ms-1
Alternative using Derived Formula
v = ? u = 25ms h = 4.0m g = 10ms-1
v = 𝑢 2 +2𝑔ℎ
v = ×10×4
v =
v  27ms-1 
38o v
126 ms-1 v
tan  = 𝒐𝒑𝒑 𝒂𝒅𝒋
tan  = 𝟖.𝟗𝟒𝟒𝟐𝟕 𝟐𝟓
 = tan-1 𝟖.𝟗𝟒𝟒𝟐𝟕 𝟐𝟓
 = o
  19o
In y direction
vy = ? uy = 0 x=4.0m a = 10ms-2
v2 = u2 + 2ax
v2 = 2 × 10 × 4
v2 =80
v =
tan  = 𝒐𝒑𝒑 𝒂𝒅𝒋
tan  = 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = tan-1 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = o
  38o
19o
27 ms-1

45. Same as derived formula
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.
Find final velocity
h2 = a2 + b2
v2 =
v2 = 705
v =
v  27 ms-1
25ms-1 
ms-1
Alternative using energy approach
v = ? u = 25ms h = 4.0m g = 10Nkg-1
Eki + Ug  Ekf
½mu2 + mgh = ½mv2
u2 + 2gh = v2
v = 𝑢 2 +2𝑔ℎ 
38o v
tan  = 𝒐𝒑𝒑 𝒂𝒅𝒋
tan  = 𝟖.𝟗𝟒𝟒𝟐𝟕 𝟐𝟓
 = tan-1 𝟖.𝟗𝟒𝟒𝟐𝟕 𝟐𝟓
 = o
  19o
126 ms-1 v
In y direction
vy = ? uy = 0 x=4.0m a = 10ms-2
v2 = u2 + 2ax
v2 = 2 × 10 × 4
v2 =80
v =
v = ×10×4
v =
v  27ms-1
tan  = 𝒐𝒑𝒑 𝒂𝒅𝒋
tan  = 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = tan-1 𝟕𝟕.𝟒𝟓𝟗𝟕 𝟏𝟎𝟎
 = o
  38o
19o
Same as derived formula
27 ms-1

46. 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
v = ? t = 1.5s u = 0 a = 𝑭 𝒏𝒆𝒕 𝒎 = 𝟐𝟐 𝟎.𝟓 = 44ms-2
v = u + at
v = × 1.5
v = 66 ms-1
In y direction
v = ? t = 1.5s u = 0 a = 10ms-2
v = × 1.5
v = 15 ms-1
22N y x
alternative using momentum/impulse approach
v = ? t = 1.5s u = Fnet = 22N
Ft = m(v – u)
22 × 1.5 = 0.5v
33 = 0.5v
66 ms-1 = v

47.  v x y 13o 68 ms-1 66ms-1 15ms-1 2006 Exam Q8 Q3 Find final velocity
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
h2 = a2 + b2
v2 =
v2 = 4581
v =
v  68 ms-1
66ms-1 
15ms-1
22N v
tan  = 𝒐𝒑𝒑 𝒂𝒅𝒋
tan  = 𝟏𝟓 𝟔𝟔
 = tan-1 𝟏𝟓 𝟔𝟔
 = o
  13o y x
13o
68 ms-1

48. Obliquely Launched Projectile Motion

49. Obliquely Launched Projectile Motion
Case 1: landing height = launch height Case 2: landing height different to launch height

50. Case 1: Obliquely Launched Projectiles Landing At The Launching Height
In y direction
uy = u sin a = –10ms-1 use const accel equns
In x dirn
vx = u cos in d v t
The motion in the x & y directions are independent
uy = u sin  vx is constant = u cos 
If not enough info in one dirn get time from the other dirn
Reference
axes
If time not involved
Eki  Ekf + Ug
½mu2 = ½mv2+ mgh
u2 = v2+ 2gh y x vx vx vy vx vy u
Ekmin = ½ mvx2
uy = u sin  vx 
ux = u cos 
=
Speeds before hmax are the same as those after at each height vy
Eight points
v = u

51. Case 1: Obliquely Launched Projectiles Landing At The Launching Height
hmax= 𝑢 2 𝑠𝑖𝑛 2 𝜃 2𝑔
range = 𝑢 2 𝑠𝑖𝑛2𝜃 𝑔
tflight = 2𝑢𝑠𝑖𝑛𝜃 𝑔 y x vx vx vy vx vy u
uy = u sin   vx
ux = u cos 
= vy
v = u
Three formulae

52. 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 horizontal distance (using t = 𝑑 𝑣 ) x u

53. Obliquely Launched Projectile Motion
Worked Examples

54. Obliquely Launched Projectile Motion Example 1
What are the three other ways of working out the flight time?
t = ? u = x =0 a = –10ms-2 (this will involve a factorisation)
t = ? u = v = – a = –10ms-2 (this is pretty easy)
Using flight time derived formula (just put the numbers in and work out)
25o
31ms-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? In x direction d = ? v = ms-1 t = ?
uy = 31sin25
= ms-1
ux = 31cos25
= ms-1 d
Find t to top of path from y dirn
t = ? u = v =0 a = –10ms-2
v = u + at
0 = – 10 × t
– = – 10 × t
= t
Find total time
Total time = 2 ×
= s

55. Obliquely Launched Projectile Motion Example 1
What are the three other ways of working out the flight time?
t = ? u = x =0 a = –10ms-2 (this will involve a factorisation)
t = ? u = v = – a = –10ms-2 (this is pretty easy)
Using flight time derived formula (just put the numbers in and work out)
uy = 31sin25
= ms-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? In x direction d = ? v = ms-1 t = ?
ux = 31cos25
= ms-1
Find t from y dirn
t = ? u = x = a = –10ms-2
x = ut + ½at2
0 = t + ½ × – 10 × t2
0 = t – – 5 t2
0 = t ( – 5t )
t = 0, 𝟏𝟑.𝟏𝟎𝟏 𝟓
t =

56. Obliquely Launched Projectile Motion Example 1
What are the three other ways of working out the flight time?
t = ? u = x =0 a = –10ms-2 (this will involve a factorisation)
t = ? u = v = – a = –10ms-2 (this is pretty easy)
Using flight time derived formula (just put the numbers in and work out)
uy = 31sin25
= ms-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? In x direction d = ? v = ms-1 t = ?
ux = 31cos25
= ms-1
Find t from y dirn
t = ? u = v = a = –10ms-2
v = u + at
= – 10 × t
– = – 10 × t
= t

57. Obliquely Launched Projectile Motion Example 1
What are the three other ways of working out the flight time?
t = ? u = x =0 a = –10ms-2 (this will involve a factorisation)
t = ? u = v = – a = –10ms-2 (this is pretty easy)
Using flight time derived formula (just put the numbers in and work out)
uy = 31sin25
= ms-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? In x direction d = ? v = ms-1 t = ? d = v t d = × d = d  74m
ux = 31cos25
= ms-1
Find t from flight formula
t = ? u = 31ms-1  = 25o g = –10ms-2
tflight = 2𝑢𝑠𝑖𝑛𝜃 𝑔
tflight = 2×31×𝑠𝑖𝑛25 10
tflight =
2.6202s

58. Obliquely Launched Projectile Motion Example 1
31ms-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? In x direction d = ? v = ms-1 t = ? d = v t d = × d = d  74m
range
1 step alternative using derived formulae
Robliq = ? u = 31ms  = 25o g = 10ms-1
Robliq = 𝑢 2 𝑠𝑖𝑛2𝜃 𝑔
Robliq = sin⁡(2×25) 10
Robliq =
Robliq  74m
Find t to top of path from y dirn
t = ? u = v =0 a = –10ms-2
v = u + at
0 = – 10 ×
– = – 10 × t
= t
Find total time
Total time = 2 ×
= s
2.6202s

59. Obliquely Launched Projectile Motion Example 1
31ms-1
A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal. (b) What is the minimum speed of the golf ball? Minimum speed occurs at the top of the path when there only the horizontal component of velocity. So minimum is ms-1
uy = 31sin25
= ms-1
ux = 31cos25
= ms-1

60. Obliquely Launched Projectile Motion Example 1
31ms-1
A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal. (c) How high will the golf ball rise?
uy = 31sin25
= ms-1
ux = 31cos25
= ms-1
In y direction
x = ? u = ms v = a = –10ms-2
v2 = u2 + 2ax
0 = × – 10 × x
0 = – 20 × x
– = – 20 × x
= x
x = 8.6m

61. Obliquely Launched Projectile Motion Example 1
31ms-1
A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal. (c) How high will the golf ball rise?
uy = 31sin25
= ms-1
ux = 31cos25
= ms-1
Alternative using derived formulae
hmax= ? u = 31ms  = 25o g = 10ms-1
hmax= 𝑢 2 𝑠𝑖𝑛 2 𝜃 2𝑔
hmax= 𝑠𝑖𝑛 ×10
hmax =8.5820
hmax  8.6m
In y direction
x = ? u = ms v = a = –10ms-2
v2 = u2 + 2ax
0 = × – 10 × x
0 = – 20 × x
– = – 20 × x
= x
x = 8.6m

62. Obliquely Launched Projectile Motion Example 1
31ms-1
A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal. (d) What is the velocity of the ball as it hits the ground? Since the ball hits the ground at the same height as it launch height final speed is:
uy = 31sin25
= ms-1
ux = 31cos25
= ms-1
25o
31ms-1

63. Obliquely Launched Projectile Motion Example 1
31ms-1
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?
Using energy approach (which avoids having to work with vectors)
v = ? u = 31ms h = 20m g = 10ms-1
Eki  Ekf + Ug
½mu2 = ½mv2 + mgh
½u2 = ½v2+ gh
u2 = v2+ 2gh
312 = v2+ 2×10×20
961 = v
561 = v2
= v
v  24 ms-1

64. Obliquely Launched Projectile Motion Example 2
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?
In y direction
x = ? u = ms-1 t = 6.0s a = –10ms-2
x = ut + ½ at2
x = × 6 + ½ × – 10 × 62
x = –
so the cliff is  94m high
35o
25ms-1
uy = 25sin35
= ms-1
ux = 25cos35
= ms-1

65. Obliquely Launched Projectile Motion Example 2
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 = ms-1 t = 6
d = v t
d = × 6
d =
d  123m d
35o
25ms-1
uy = 25sin35
= ms-1
ux = 25cos35
= ms-1

66. Obliquely Launched Projectile Motion Example 3
28ms-1 h
A golf ball is hit at 28ms-1 at 60o 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?
In y direction
x = ? u = ms-1 a = –10ms-2 t = ?
x = ut + ½ at2
x = × 4 + ½ × – 10 × 42
x =
so the ball hits the tree  17m up
60o
60o
28ms-1
56m
uy = 28sin60
= ms-1
ux = 28cos60
= 14ms-1
Find t from the x direction
t = ? v = ux = 14ms d= 56m
t = 𝒅 𝒗
t = 𝟓𝟔 𝟏𝟒
t = 4 s 4s

67. Obliquely Launched Projectile Motion
Exam Questions

68. 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 =  = 20o g = 10ms-2
Robliq = 𝑢 2 𝑠𝑖𝑛2𝜃 𝑔
10 = 𝑢 2 sin⁡(2×20) 10
100 = u2sin40
= u2
= u
u  12 ms-1
Using x & y direction analysis for this problem is very difficult so using the range formula is a much more practical way to solve this problem

69. u y x 30ms-1 8o 3.0m 30ms-1 cos  = 𝑎𝑑𝑗 ℎ𝑦𝑝 cos 8 = 30 𝑢 u = 30 𝑐𝑜𝑠8
2005 Exam Q11 Q2 y x 8o
3.0m
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  = 𝑎𝑑𝑗 ℎ𝑦𝑝
cos 8 = 30 𝑢
u = 𝑐𝑜𝑠8
u = ms-1
u  30.3ms-1 u 8o
30ms-1

70. 30ms-1
2005 Exam Q12 Q3 y x x 8o
3.0m
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 = ? u = 30ms-1 v = 0 a = –10ms-2 v2 = u2 + 2ax 0 = × – 10 × x 0 = – 20 × x – = – 20 × x = x Overall height = =  3.89 m
ms-1
sin8
= ms-1 8o
30ms-1

71. y x 30ms-1 hmax 8o 3.0m 30ms-1 3.0m Alternative using derived formulae
2005 Exam Q12 Q3 y x
hmax 8o
3.0m
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 = ? u = 30ms-1 v = 0 a = –10ms-2 v2 = u2 + 2ax 0 = × – 10 × x 0 = – 20 × x – = – 20 × x = x Overall height = =  3.89 m
Alternative using derived formulae
hmax= ? u = ms  = 8o g = 10ms-2
hmax= 𝑢 2 𝑠𝑖𝑛 2 𝜃 2𝑔
hmax= 𝑠𝑖𝑛 2 8 2×10
hmax =
ms-1 8o
30ms-1

72. y x 30ms-1 8o h2 h1 = 3.0m 30ms-1 Alternative using Energy Approach
2005 Exam Q12 Q3 y x 8o h2
h1 =
3.0m
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. At its highest point, how far was the ball above the ground. x = ? u = 30ms-1 v = 0 a = –10ms-2 v2 = u2 + 2ax 0 = × – 10 × x 0 = – 20 × x – = – 20 × x = x Overall height = =  3.89 m
Alternative using Energy Approach
h2= ? u = ms h1 = 3m g = 10ms-2
Eki + Ugi  Ekf + Ugf
½ mu2 + mgh1 = ½ mv2 + mgh2
½ u2 + gh1 = ½ v2 + gh2
½ × × 3 = ½ × × h2
= × h2
= h2
h2  3.89m
ms-1 8o
30ms-1

73. y x 40ms-1 uy = 40sin25 = 16.9047ms-1 25o ux = 40cos25 = 36.2523ms-1
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?
25o
40ms-1 y x
Find t from the x direction
t = ? v = ux = ms-1 d= 127m
t = 𝒅 𝒗
t = 𝟏𝟐𝟕 𝟐𝟖.𝟎𝟗𝟔
t = s
t  3.5 s
uy = 40sin25
= ms-1
ux = 40cos25
= ms-1

74. Alternative y x 40ms-1 uy = 40sin25 = 16.9047ms-1 25o ux = 40cos25
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
Find t from the y direction
t = ? u = ms-1 v = – ms-1 a = –10ms-2
v = u + at
= – 10 × t
– = – 10 × t
= t
t  3.4 s
25o
40ms-1 y x
uy = 40sin25
= ms-1
ux = 40cos25
= ms-1
There is a discrepancy in the data for this problem hence the slight difference in answers from the x direction and y direction

75. y x 40ms-1 uy = 40sin25 = 16.9047ms-1 25o ux = 40cos25 = 36.2523ms-1
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?
Find t from the y direction
t = ? u = ms-1 v = – ms-1 a = –10ms-2
v = u + at
= – 10 × t
– = – 10 × t
= t
t  3.4 s
25o
40ms-1 y x
uy = 40sin25
= ms-1
ux = 40cos25
= ms-1
There is a discrepancy in the data for this problem hence the slight difference in answers from the x direction and y direction

76. Alternative with formula
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
Find t from flight formula
t = ? u = 40ms-1  = 25o g = –10ms-2
tflight = 2𝑢𝑠𝑖𝑛𝜃 𝑔
tflight = 2×40×𝑠𝑖𝑛25 10
tflight =
tflight  3.4s
25o
40ms-1 y x
uy = 40sin25
= ms-1
ux = 40cos25
= ms-1
There is a discrepancy in the data for this problem hence the slight difference in answers from the previous methods

77. y x 40ms-1 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?
25o
40ms-1 y x
Find hmax from y direction
x = ? u = ms-1 v = a = –10ms-2
v2 = u2 + 2ax
0 = × – 10 × x
0 = – 20 × x
– = – 20 × x
= x
x  14m
uy = 40sin25
= ms-1
ux = 40cos25
= ms-1

78. Alternative with formula
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
Find t from flight formula
hmax= ? u = 40ms  = 25o g = 10ms-2
hmax= 𝑢 2 𝑠𝑖𝑛 2 𝜃 2𝑔
hmax= 𝑠𝑖𝑛 ×10
hmax =
hmax  14m
25o
40ms-1 y x
uy = 40sin25
= ms-1
ux = 40cos25
= ms-1
There is a discrepancy in the data for this problem hence the slight difference in answers from the previous method

79. 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.

80. 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 B. initial speed C. acceleration D. angle of firing level?
Because the paintball gets to the same vertical height the vertical component in each situation must be the same and hence the flight time will also be the same.
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.

81. Some Loose Rules To Follow.
So With More Than One Method For Solving Most Projectile Motion Questions – How Do I Decide On The Best?
Some Loose Rules To Follow.

82. Some Loose Rules For Deciding On The Best Method To Solve A Problem
The best method is the one that makes the most sense to you and doesn’t take forever to get to an answer
Often if there is a formula use a formula, but be careful of expressions such as
hmax= 𝑠𝑖𝑛 2 8 2×10 =
Robliq = 𝑢 2 𝑠𝑖𝑛2𝜃 𝑔
v = ×10×300 =
Rhproj = × =
Make sure you practice more than one method when solving projectile motion questions – particularly working with x & y directions.

83. Some Loose Rules For Deciding On The Best Method To Solve A Problem
An energy approach is very useful when asked for the magnitude of a velocity that is not at launching height or maximum height.
Remember there are three approaches that can be used for solving
Constant Accel Equations
Impulse/ Momentum Formulae
(no dist)
Energy Formulae
(no time)