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Velocity at any point of projectile
Intermediate
Horizontal component is constant; vertical changes at g.
v = √(vx²+vy²)At any time vx = u cosθ and vy = u sinθ − gt. The speed is √(vx² + vy²) and the direction tanφ = vy/vx. Energy conservation gives speed at height y: v² = u² − 2gy.
At height 10 m launched at 20 m/s: v = √(400 − 196) ≈ 14.3 m/s.
Speed at equal heights is the same on the way up and down.
In more detail
At any point on a projectile's path, the velocity is tangent to the trajectory. Its horizontal component stays constant, while its vertical component changes at g.
Real-world uses
- Predicting the impact speed of a dropped or thrown object at the ground.
- Calculating the direction a ball travels when it lands.
- At the top of a jump, the velocity is purely horizontal, so the speed there is the horizontal component alone.
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Frequently asked questions
What is the velocity at the highest point?
Purely horizontal, since the vertical component is zero there.
How do I find the speed at a given point?
Combine the horizontal and vertical components using Pythagoras.
What is the speed at the top of a projectile path?
It equals the horizontal component of the launch velocity, because the vertical component is zero at the peak.
Related topics in Kinematics
Horizontal projectile · Projectile on an incline · Trajectory equation · Effect of air resistance · Relative velocity in 2D · River-boat problems