1. Unit 1 Our Dynamic Universe Projectiles
CfE Higher Physics
Unit 1
Our Dynamic Universe
Projectiles

2. Learning Intentions
I can carry out calculations to find the horizontal and vertical components of vectors using the relationships:
VH = VcosӨ Vv = VsinӨ

3. Projectiles
A projectile is any object that has been thrown/fired and is under the influence of a gravitational field. The only force acting on a projectile is its weight acting vertically downwards. A projectile therefore accelerates in the downward direction.
W = mg
Projectiles follow a curved path.

4. Solving Numerical Problems
The horizontal and vertical motions of a projectile should be considered separately.
Horizontally - constant velocity (no forces) (v=d/t only)
Vertically - constant downward acceleration (due to weight) (SUVAT only)
For the horizontal motion we can use the displacement, velocity, time formula (velocity is constant).

5. Solving Numerical Problems
For the vertical motion we must use equations of motion (constant acceleration).
The horizontal and vertical motion are linked by time.

6. Hints and Tips: Acceleration due to gravity is always negative.
If the object travels down, displacement is negative.
Time taken to travel horizontally is the same vertically.
The time to reach its max height is half the full time of flight.
Split into horizontal and vertical motions.
When you use VcosӨ or VsinӨ you are calculating the initial velocity horizontally or vertically!
The vertical velocity at the max height is 0 ms-1

7. v OR Projectiles t vv = v sin θ θ vH = v cos θ d = v x t E.O.M.
Horizontal
Vertical
d = v x t
E.O.M.
d = ?
v =v cos θ
t = ?
s = ?
u = v sin θ
v = ?
a = m s-2 t

8. More Rules on Projectiles…
t= ½ total time
v= 0 m/s
At maximum height, vertical speed is zero (v=0)
At maximum height, the horizontal distance is half the final value
At maximum height, the time taken is half the final value θ d1 d2

9. Worked Example 1
A golf ball leaves the tee with a velocity of 42 ms-1 at an angle of 35° to the ground. How far does the ball travel before it hits the ground?
Initial horizontal velocity = 42cos35°
uh = 34.4 ms-1
42 ms-1
35°
Initial vertical velocity = 42sin35°
uv = 24.1 ms-1

10. Worked Example 1 (continued)
Vertical motion:
sv = uvt + ½at2
sv =
0 m
uv =
24.1 ms-1
0 = 24.1t + ½(-9.8)t2
vv =
0 = 24.1t - 4.9t2
a =
-9.8 ms-2
0 = t(24.1 – 4.9t)
t = ?
t = 0 s or s
upwards is positive
leaves tee
returns to ground

11. Worked Example 1 (continued)
Horizontal motion:
sh = vht
= 34.4 x 4.92
= 169 m

12. Worked Example 2
A car travelling at 16 ms-1 drives horizontally off the edge of a 62 m cliff. With what velocity does the car hit the water?
Vertical motion:
vv2 = uv2 + 2asv
sv =
-62 m
uv =
0 ms-1
vv2 = (-9.8)(-62)
vv = ?
vv2 = 1215
a =
9.8 ms-2
t =
vv = 34.9 ms-1

13. Worked Example 2 (continued)
Horizontal motion:
vh = 16 ms-1
Combining the two:
magnitude = √( )
= 38.4 ms-1
direction, θ = tan-1(34.9/16)
= 65.4° to horizontal
16 ms-1 θ
34.9 ms-1

14. Exam Style Questions – stick into notes