Advanced · 14 min
Vectors and projectile motion
Resolve motion into perpendicular components and predict a horizontal launch.
Components describe direction
A vector has magnitude and direction. Perpendicular components let you work along two axes separately. A velocity of magnitude v at angle θ above the horizontal has components vx = v × cos θ and vy = v × sin θ. Its magnitude can be recovered from √(vx² + vy²). Components are signed: changing direction changes the sign on the relevant axis.
Worked example
A velocity has horizontal component 3 m/s and vertical component 4 m/s. Speed = √(3² + 4²) = 5 m/s
One motion, two calculations
For a projectile with no air resistance and constant downward gravity, horizontal velocity stays constant while vertical velocity changes. A horizontal launch begins with zero vertical velocity. If its launch height is h, its time to fall is found from h = ½ × g × t². The same time applies horizontally, so horizontal distance is vx × t. Use this model only where gravity can be treated as uniform.
Worked example
A ball launches horizontally at 4 m/s from a height of 5 m, with g = 10 m/s². t² = 2 × 5 ÷ 10 = 1; time = 1 s Horizontal distance = 4 × 1 = 4 m
Separate the two axes
- For a horizontal launch at 3 m/s from 5 m high, find the fall time and horizontal distance using g = 10 m/s².
- Repeat with horizontal speed 6 m/s. Explain why the horizontal distance changes but the fall time does not in this model.