Projectile Motion and Free Fall
Projectile Motion and Free Fall
Projectile motion is the motion of an object under the influence of gravity alone, after being launched with some initial velocity. Air resistance is neglected in standard A-Level analysis.
Key Principle: Independence of Horizontal and Vertical Motion
The horizontal and vertical components of a projectile's motion are completely independent of each other. This is the fundamental principle that makes projectile problems solvable.
- Horizontal motion: constant velocity (no acceleration, assuming no air resistance)
- Vertical motion: constant acceleration due to gravity, g = 9.81 m s⁻² downward
Free Fall
Free fall is the special case where an object moves under gravity alone with no horizontal velocity. The only force acting is weight.
Key equations for free fall (taking downward as positive):
| Equation | Variables |
|---|---|
| v = u + gt | v = final velocity, u = initial velocity, g = 9.81 m s⁻², t = time |
| s = ut + ½gt² | s = displacement |
| v² = u² + 2gs | No time needed |
| s = ½(u + v)t | Average velocity form |
Worked example — dropping a ball: A ball is dropped from rest from a height of 20 m. Find the time to reach the ground and the impact speed.
Taking downward as positive, u = 0, s = 20 m, g = 9.81 m s⁻²:
s = ut + ½gt² → 20 = 0 + ½(9.81)t² → t² = 40/9.81 = 4.077 → t = 2.02 s
v = u + gt = 0 + 9.81 × 2.02 = 19.8 m s⁻¹
Resolving Projectile Motion
For an object launched at angle θ to the horizontal with initial speed u:
- Horizontal component: uₓ = u cos θ
- Vertical component: uᵧ = u sin θ
Horizontal displacement at time t: x = (u cos θ)t
Vertical displacement at time t: y = (u sin θ)t − ½gt²
Finding the Range
The range is the total horizontal distance travelled when the projectile returns to its launch height. Setting y = 0:
0 = (u sin θ)t − ½gt² → t(u sin θ − ½gt) = 0
So t = 0 (launch) or t = 2u sin θ / g (landing).
Substituting into x: Range R = u² sin 2θ / g
This shows the maximum range occurs at θ = 45° (since sin 2θ is maximised when 2θ = 90°).
Finding Maximum Height
At maximum height, the vertical component of velocity is zero. Using v² = u² + 2as vertically:
0 = (u sin θ)² − 2g·H → H = u² sin²θ / (2g)
Worked Example — Full Projectile Calculation
A football is kicked at 18 m s⁻¹ at 35° above the horizontal. Find:
(a) Time of flight:
uᵧ = 18 sin 35° = 10.32 m s⁻¹
t = 2uᵧ/g = 2 × 10.32 / 9.81 = 2.10 s
(b) Range:
uₓ = 18 cos 35° = 14.74 m s⁻¹
R = uₓ × t = 14.74 × 2.10 = 31.0 m
(c) Maximum height:
H = uᵧ² / (2g) = 10.32² / (2 × 9.81) = 5.43 m
Projectiles Launched Horizontally
When launched horizontally from a height h, the initial vertical velocity is zero:
- Horizontal: x = ut (where u is the launch speed)
- Vertical: h = ½gt² → t = √(2h/g)
The velocity at any instant is found by combining components: v = √(vₓ² + vᵧ²)
The angle below the horizontal: tan α = vᵧ / vₓ = gt / u
The Effect of Air Resistance (Qualitative)
In reality, air resistance:
- Reduces the range and maximum height
- Makes the trajectory asymmetric — the descent is steeper than the ascent
- Means the landing speed is less than the launch speed
- Eventually produces a terminal velocity if the fall is long enough
Terminal Velocity
During free fall through a fluid, drag force increases with speed. When drag force equals weight, the resultant force is zero and the object reaches terminal velocity.
Weight = Drag → mg = F_drag
At terminal velocity, acceleration = 0, and the object falls at constant speed. The value depends on mass, cross-sectional area, and the drag coefficient.
Experimental Determination of g
Method: Drop a ball bearing through two light gates a measured distance apart.
1. Measure the distance s between the gates
2. Record the time t for the ball to travel between them
3. The ball's speed at the first gate u can be found from the light gate
4. Use s = ut + ½gt² to calculate g
5. Repeat and average to reduce random error
Systematic errors: air resistance (gives g too low), timing delays, parallax in measuring s.