Horizontal Velocity Component
Worked example: 20 m/s at 60° → 10 m/s. Press Try an example to run it live, then adjust anything.
Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Change any unit and the answer follows.
Where Horizontal Velocity Component fits
- The projectile motion equations6 formulas
See also
Projectile Maximum Height: Both use launch speed and launch angle.Projectile Range on Level Ground: Both use launch speed and launch angle.Projectile Time of Flight: Both use launch speed and launch angle.Vertical Velocity Component: Both use launch speed and launch angle.Horizontal Velocity Component explained
Every projectile problem starts by splitting the launch velocity into two independent pieces, and the horizontal one is v cos θ. With no air resistance nothing acts sideways, so vₓ never changes from launch to landing: the projectile drifts downrange at a fixed rate while gravity works only on the vertical component. Throw at 20 m/s and 60° above horizontal and vₓ = 20 × cos 60° = 10 m/s for the entire flight.
The classic trap is swapping sine and cosine. Anchor it with the extremes: at θ = 0° the throw is entirely horizontal and cos 0° = 1 keeps the whole speed, while at θ = 90° the throw is straight up and cos 90° = 0 leaves nothing moving sideways. Multiply vₓ by the time of flight and you have the range, which is exactly how the range formula is derived.
Horizontal Velocity Component formula
- = Horizontal velocity (m/s)
- = Launch speed (m/s)
- = Launch angle (°)
Cite this page
formula.expert. (2026, October 16). Horizontal Velocity Component Formula & Calculator. https://formula.expert/solvers/projectile-horizontal-velocity-component
https://formula.expert/solvers/projectile-horizontal-velocity-component