1. Projectile Motion Modelling assumptions
Objects thrown in a PROJECTILE MOTION fly in a parabola
Modelling assumptions
a projectile is a particle (negligible mass)
no air resistance
no wind
not powered

2. This is no good.
Wind and air resistance will effect the motion too much.

3. Freddie hitting a six

4. Velocity An object is projected at 40ms-1 at 50o to the x-axis
“a measure of speed and direction”
An object is projected at 40ms-1 at 50o to the x-axis N
What are the horizontal and vertical components
of the velocity?
40ms-1
40 sin 50o
50o
40 cos 50o

5. “Resolving Velocities”
E.g a ball thrown with velocity U at degrees to the horizontal
U ms-1
U sin  
U cos 
Horizontal component = U cos 
Vertical component = U sin 

6. Projectile Problems 15 ms-1
“A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal.
Find the:-
time of flight
the range
the maximum height “
15 ms-1
30o O x y
Maximum
Height
Range

7. Equations
Horizontal motion
Vertical Motion
Initial position a u v r

8. Equations Initial position a -g u v r Horizontal motion
Vertical Motion
Initial position a -g u v r

9. Equations Initial position a -g u ux = 15cos30 uy=15sin30 v r
Horizontal motion
Vertical Motion
Initial position a -g u
ux = 15cos30
uy=15sin30 v r

10. Equations Initial position a -g u ux = 15cos30 uy=15sin30 v (=u+at)
Horizontal motion
Vertical Motion
Initial position a -g u
ux = 15cos30
uy=15sin30
v (=u+at)
vx =
vy = r

11. Equations Initial position a -g u ux = 15cos30 uy=15sin30 v (=u+at)
Horizontal motion
Vertical Motion
Initial position a -g u
ux = 15cos30
uy=15sin30
v (=u+at)
vx = 15cos30
vy = 15cos30-gt r

12. Equations Initial position a -g u ux = 15cos30 uy= 15sin30 v (=u+at)
Horizontal motion
Vertical Motion
Initial position a -g u
ux = 15cos30
uy= 15sin30
v (=u+at)
vx = 15cos30
vy = 15sin30-gt
r (=ut+0.5at2)
rx=
ry =

13. Equations Initial position a -g u ux = 15cos30 uy= 15sin30 v (=u+at)
Horizontal motion
Vertical Motion
Initial position a -g u
ux = 15cos30
uy= 15sin30
v (=u+at)
vx = 15cos30
vy = 15sin30-gt
r (=ut+0.5at2)
rx=15cos30t
ry = 15sin30t-.5gt2

14. Find the time of flight

15. Find the time of flight When flight is complete y=0
ry = 15sin30t-.5gt2 = 0

16. Find the time of flight When flight is complete y=0
ry = 15sin30t-.5gt2 = 0
0 = 15 sin 30 t + 1/2 x -9.8 x t2
Either: t = 0 [start]
or t = 0
t = 7.5/4.9
= 1.53 sec.
0 = 7.5t - 4.9t2
0 = t( t)

17. Find the range
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range

18. Find the range
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range.
The range is the horizontal distance travelled before landing.

19. Find the range
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range.
The range is the horizontal distance travelled before landing.
This is the value of x when y=0.

20. Find the range
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range.
The range is the horizontal distance travelled before landing.
This is the value of x when y=0.
rx=15cos30t

21. Find the range rx = 15 cos 30 x 1.53 rx = 12.99 x 1.53 = 19.9 m (1 dp)
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the range.
rx = 15 cos 30 x 1.53
rx = x 1.53 = 19.9 m (1 dp)

22. Find the maximum height
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height

23. Find the maximum height
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height
Max height occurs when the vertical velocity is zero.
This happens at the top of the flight.

24. Find the maximum height
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height
Max height occurs when the vertical velocity is zero.
This happens at the top of the flight.
vy = 15sin30-gt
0 = t
0 = 15 sin t
t = 7.5 / 9.8 = sec

25. Find the maximum height
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height
Max height occurs when the vertical velocity is zero.
This happens at the top of the flight.
vy = 15sin30-gt
0 = t
0 = 15 sin t
t = 7.5 / 9.8 = sec
We now need the value of y at this time

26. Projectile Problem - so far
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal.
Find the:-
time of flight sec
the range m
the maximum height when t = 0.765
Notice the time at the maximum height is exactly half the total time of flight
15 ms-1
30o O x y
Maximum
Height
For ‘symmetrical’ problems like this that will always be the case
Range

27. A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height
t = sec
ry = 15sin30t-.5gt2

28. t = 0.765 sec ry = 15sin30t-.5gt2 = 2.87 ~ 2.9 m (1 dp)
A particle of projected from O with initial velocity 15 ms-1 at an angle of 300 to the horizontal. Find the maximum height
t = sec
ry = 15sin30t-.5gt2
= ~ 2.9 m (1 dp)

29. Projectile Problem - with height elements
“A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal.
How long is it more than 2m above the ground ? “
12 ms-1
40o O x y
2 m

30. Equations Initial position a -g u ux = 12cos40 uy= 12sin40 v (=u+at)
Horizontal motion
Vertical Motion
Initial position a -g u
ux = 12cos40
uy= 12sin40
v (=u+at)
vx = 12cos40
vy = 12sin40-gt
r (=ut+0.5at2)
rx=12cos40t
ry = 12sin40t-.5gt2

31. A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal. How long is it more than 2m above the ground ?
ry = 12sin40t-.5gt2

32. A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal. How long is it more than 2m above the ground ?
ry = 12sin40t-.5gt2
2 = 12 sin 40 t - 1/2 x 9.8 x t2
2 = 7.713t - 4.9t2
4.9t t + 2 = 0

33. A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal. How long is it more than 2m above the ground ?
ry = 12sin40t-.5gt2
2 = 12 sin 40 t - 1/2 x 9.8 x t2
2 = 7.713t - 4.9t2
4.9t t + 2 = 0
t = 0.33 or (2 dp)

34. t = 0.33 or 1.25 (2 dp) Projectile Problem - with height elements
“A ball is hit with initial velocity 12 ms-1 at an angle of 400 to the horizontal.
How long is it more than 2m above the ground ? “
t = 0.33 or (2 dp)
It is above 2m for
= 0.92 seconds
12 ms-1
40o O x y
2 m