Projectile Range
Horizontal distance a ball travels when launched at speed v and angle θ.
Formula
Variables
Example
30 m/s at 45°: R = 900×1÷9.81 ≈ 91.7 m
How it works
This formula gives the horizontal range of a projectile launched and landing at the same height, assuming air resistance is negligible. It works because the launch speed can be split into horizontal and vertical parts: the horizontal part sets how long the object keeps moving across, while the vertical part sets how long it stays in the air. Combining those two effects leads to a maximum range at 45° for a fixed launch speed.
When to use it
Use this when analysing balls, shots, or launched objects that start and finish at the same level, such as a football kick, golf shot, or thrown ball. It is especially useful in sports science and motion problems where you know the launch speed and angle and want the horizontal distance.
Worked example, step by step
Given: v = 20 m/s, θ = 30°, g = 9.81 m/s² Step 1: Find 2θ 2θ = 2 × 30° = 60° Step 2: Find sin(2θ) sin(60°) = 0.866 Step 3: Square the speed v² = 20² = 400 Step 4: Substitute into the formula R = (v² sin(2θ)) / g R = (400 × 0.866) / 9.81 Step 5: Calculate R = 346.4 / 9.81 ≈ 35.3 m Final answer: The projectile travels about 35.3 m horizontally.
Common mistakes to avoid
- Forgetting that the angle in the formula is doubled before taking the sine
- Using degrees and radians inconsistently on a calculator
- Applying the formula when the launch and landing heights are different or when air resistance is important
Did You Know?
A 45° launch gives maximum range on flat ground — but for a shot put released above the ground the optimum is closer to 42°.
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