Projectile Motion
Analyse the motion of projectiles by resolving velocity into horizontal and vertical components, and calculate trajectory, range, and maximum height.
What is Projectile Motion?
A projectile is any object that, once launched, moves only under the influence of gravity (ignoring air resistance). The key insight is that projectile motion can be analysed as two independent motions: horizontal (constant velocity) and vertical (uniformly accelerated by gravity).
Horizontal Component
vx = v cos θ
x = vx t
No horizontal acceleration (ignoring air resistance), so the horizontal velocity remains constant throughout the flight.
Vertical Component
vy = v sin θ
y = vyt - ½gt²
Gravity acts downward at g = 9.8 m s-2, causing the vertical velocity to change constantly.
Key principle: The horizontal and vertical motions are completely independent of each other. Gravity only affects the vertical component.
Trajectory, Range and Maximum Height
When a projectile is launched at angle θ from the ground with initial speed v, its path forms a parabola. We can derive key quantities for the motion.
Projectile Trajectory
Launch Point
Velocity resolved into vx and vy
Maximum Height
vy = 0, hmax = vy² / (2g)
Landing Point
Range R = v² sin 2θ / g
Time of flight
T = 2v sin θ / g
Maximum height
H = v² sin²θ / 2g
Range
R = v² sin 2θ / g
Maximum range: For a given launch speed on level ground, the maximum range is achieved at a launch angle of 45 degrees, where sin 2θ = 1.
Special Cases of Projectile Motion
There are several common scenarios in projectile motion problems. Understanding these special cases helps simplify calculations.
Horizontal Launch (drop from a cliff)
Initial vertical velocity = 0. The object accelerates downward due to gravity while moving horizontally at constant speed. θ = 0, so vy,initial = 0.
Vertical Launch (straight up)
No horizontal component. The object goes straight up, decelerates, stops momentarily at max height, then falls back down. θ = 90 degrees.
Symmetry of Projectile Motion
Time symmetry: The time to reach maximum height equals the time to fall back to the same level.
Speed symmetry: The speed at any given height is the same going up as it is coming down (ignoring air resistance).
Complementary angles: Launch angles of θ and (90 - θ) give the same range on level ground (e.g., 30 degrees and 60 degrees).
Key Vocabulary
Projectile
An object moving under the sole influence of gravity after being launched. Air resistance is typically ignored.
Trajectory
The curved path followed by a projectile through space. For ideal projectiles, this is a parabola.
Range
The total horizontal distance travelled by a projectile from launch to landing on the same level.
Component
The resolved part of a vector in a particular direction (horizontal or vertical), found using trigonometry.
Worked Examples
A ball is kicked at 20 m s-1 at 30 degrees to the horizontal. Find the horizontal and vertical components of velocity.
Step 1: vx = v cos θ = 20 cos 30° = 20 × 0.866 = 17.3 m s-1.
Step 2: vy = v sin θ = 20 sin 30° = 20 × 0.5 = 10.0 m s-1.
Answer: The horizontal component is 17.3 m s-1 and the vertical component is 10.0 m s-1.
A stone is thrown horizontally at 15 m s-1 from a cliff 80 m high. How long does it take to reach the ground?
Step 1: Vertical motion only: s = ½gt² (initial vertical velocity = 0).
Step 2: 80 = ½ × 9.8 × t² ⇒ t² = 80 / 4.9 = 16.33.
Step 3: t = √16.33 = 4.04 s.
A projectile is launched at 40 m s-1 at 45 degrees on level ground. Find the maximum height and range.
Step 1: vy = 40 sin 45° = 28.28 m s-1. At max height, vy = 0.
Step 2: H = vy² / (2g) = (28.28)² / (2 × 9.8) = 800 / 19.6 = 40.8 m.
Step 3: R = v² sin 2θ / g = (40)² sin 90° / 9.8 = 1600 / 9.8 = 163.3 m.
Answer: Maximum height is 40.8 m and range is 163.3 m.
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
What is the horizontal acceleration of a projectile (ignoring air resistance)?
Question 2
At what angle should a projectile be launched for maximum range on level ground?
Question 3
At the maximum height of a projectile's trajectory, which statement is true?
Question 4
A ball is thrown horizontally at 10 m s-1 from a 20 m high building. What is its horizontal distance from the building when it hits the ground? (g = 9.8 m s-2)
Question 5
Two projectiles are launched with the same speed but at 30 degrees and 60 degrees. Which has the greater range on level ground?
Key Concepts Summary
- ●Projectile motion has two independent components: horizontal (constant velocity) and vertical (accelerated by gravity).
- ●Resolve the initial velocity using vx = v cos θ and vy = v sin θ.
- ●Maximum range on level ground occurs at a launch angle of 45 degrees.
- ●At maximum height, the vertical velocity is zero but the horizontal velocity is unchanged.
- ●Complementary angles (θ and 90° - θ) produce the same range for the same launch speed.