Scenario Preset:
Max Height
0.00 m
Horizontal Range
0.00 m
Time of Flight
0.00 s
Impact Speed
0.00 m/s
Impact Angle
0.00 °
Final Velocity (vᵧ)
0.00 m/s

Components of Projection

A projectile follows a parabolic path because its motion can be split into two independent vectors: a constant horizontal velocity and a vertical motion subject to constant gravitational acceleration.

Horizontal: x = u cos(θ)t
Vertical: y = u sin(θ)t - ½gt²

1. Independence of Motion

The key principle of projectile motion is that the horizontal and vertical motions do not affect each other. Gravity only acts vertically, meaning the horizontal velocity uₓ = u cos(θ) remains constant throughout the flight.

2. The Trajectory Equation

By eliminating time (t) from the horizontal and vertical equations, we get the parabolic equation for any point (x,y) in the flight path:

y = x tan(θ) - (gx²) / (2u² cos²θ)
Exam Tip: At the peak of the trajectory, the vertical velocity is exactly zero (vᵧ = 0). Use this fact to solve for the maximum height by applying v² = u² + 2as to the vertical component.

Deep Dive: Worked Examples

✅ Example 1: Basic Ground-to-Ground

A football is kicked at 20 m/s at an angle of 30°. Calculate the horizontal range.

Step 1: Components uₓ = 20 cos 30 = 17.32 m/s
uᵧ = 20 sin 30 = 10.00 m/s
Step 2: Time of Flight v = u + at → 0 = 10 - 9.81tᵣᵢₛₑ → tᵣᵢₛₑ = 1.02s
Total Time T = 2 × 1.02 = 2.04s
Step 3: Range R = uₓT = 17.32 × 2.04 = 35.3 m

✅ Example 2: Velocity at a Certain Time

A projectle is fired at 25 m/s at 40°. Find its velocity vector after 1.5 seconds.

Step 1: Calculate Components at t=1.5s vₓ = 25 cos 40 = 19.15 m/s (constant)
vᵧ = 25 sin 40 - (9.81)(1.5) = 16.07 - 14.72 = 1.35 m/s
Step 2: Find Magnitude V = √(vₓ² + vᵧ²) = √(19.15² + 1.35²) = 19.2 m/s

✅ Example 3: Horizontal Launch

A stone is thrown horizontally from a cliff 45m high at 15 m/s. How far from the base does it land?

Step 1: Vertical Time s = ut + ½at² → 45 = 0 + ½(9.81)t²
t = √(90 / 9.81) = 3.03s
Step 2: Horizontal Distance x = vₓt = 15 × 3.03 = 45.5 m

✅ Example 4: Uneven Ground

A golf ball is hit from a tee 5m above the fairway. u = 30 m/s, θ = 20°. Find the total horizontal distance.

Step 1: Vertical components uᵧ = 30 sin 20 = 10.26 m/s
Step 2: Quadratic for Time -5 = 10.26t - 4.905t² → 4.905t² - 10.26t - 5 = 0
Solving quadratic: t ≈ 2.50s
Step 3: Distance x = (30 cos 20) × 2.50 = 28.19 × 2.50 = 70.5 m

✅ Example 5: Angle for Specified Range

At what angle should a ball be thrown at 15 m/s to hit a target 20m away on ground level?

Step 1: Range Formula R = (u² sin 2θ) / g
Step 2: Rearrange for θ sin 2θ = Rg / u² = (20 × 9.81) / 15² = 196.2 / 225 = 0.872
Step 3: Solve 2θ = arcsin(0.872) = 60.7° → θ = 30.4° (or 59.6°)