Projectile Range
Horizontal distance of a projectile on level ground.
Formula
Variables
Example
Max range at 45°
How it works
This formula gives the horizontal distance a projectile travels before it lands back at the same height it was launched from, assuming air resistance is negligible. It works because the motion can be split into two independent parts: constant horizontal speed and vertical motion under gravity. The sin 2 term appears when you combine the launch speed into horizontal and vertical components and eliminate the time of flight.
When to use it
Use it for idealized projectile-motion problems on level ground, such as balls, stones, or launched objects that start and finish at the same height. It is especially useful in physics problems asking for range, launch angle, or how changing speed affects distance.
Worked example, step by step
A football is kicked at 20 m/s at an angle of 30 to the horizontal. Find the range. R = (v^2 sin 2)/g R = (20^2 sin 60 )/9.8 R = (400 0.866)/9.8 R = 346.4/9.8 R 35.3 m So the football travels about 35.3 metres horizontally before landing.
Common mistakes to avoid
- Using the formula when the launch and landing heights are different
- Forgetting to convert the angle into 2 before taking the sine
- Using g = 10 m/s^2 without checking whether the question expects 9.8 m/s^2
Did You Know?
A projectile flies farthest when launched at 45°, and complementary angles like 30° and 60° give equal ranges.
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