Maximum Height
Simple Explanation
The maximum height is the highest point a projectile reaches, where its vertical velocity has momentarily dropped to zero before gravity pulls it back down.
Why Do We Need It?
Maximum height is a key measure of a projectile's trajectory, used in contexts from sports (how high a ball goes) to engineering (clearance needed over an obstacle).
Formula
Maximum Height (Projectile Motion)
H = u² sin²θ / 2g
The highest vertical point a projectile reaches, where its vertical velocity momentarily becomes zero.
- H
- — Maximum height, in metres
- u
- — Initial launch speed, in m/s
- θ
- — Launch angle above the horizontal
- g
- — Acceleration due to gravity, 9.8 m/s²
When to use it: Use to find the peak height of a projectile's trajectory.
Worked Example
Finding maximum height
A javelin is thrown at 22 m/s at 35° above horizontal. Find its maximum height (g = 9.8 m/s², sin35° ≈ 0.574).
Why Does This Work?
At maximum height, the vertical velocity component is momentarily zero — this formula comes from applying the equation v² = u² − 2gh to just the vertical component (initial vertical speed u sinθ) and solving for h when v = 0.
Real-Life Example
A high jumper's centre of mass
A high jumper launches their body up and over the bar at an angle.
The maximum height formula (applied to their launch speed and angle) predicts how high their centre of mass rises — athletes and coaches use this relationship to understand how launch speed and angle affect jump height.
Practice
A ball is launched straight up (θ = 90°) at 14 m/s. Find its maximum height (g = 9.8 m/s²).
MediumCommon mistake
Using the full launch speed u instead of its vertical component (u sinθ) — the formula already accounts for this via the sin²θ term, so don't apply sinθ a second time separately.
Quick Review
- H = u² sin²θ / 2g gives the peak height of a projectile.
- Maximum height occurs when vertical velocity momentarily equals zero.
- A steeper launch angle (closer to 90°) gives greater height for the same speed.