The formula
- Velocity components at launch
v₀ₓ = v₀·cos θ₀ and v₀ᵧ = v₀·sin θ₀- Horizontal motion, with zero acceleration
x = x₀ + v₀ₓt- Vertical motion, under gravity
y = y₀ + v₀ᵧt - ½gt² and vᵧ = v₀ᵧ - gt- Range, time of flight and peak height on level ground
R = v₀²·sin 2θ₀ / g, t = 2v₀·sin θ₀ / g, h = v₀ᵧ² / 2g
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
v₀ | Launch speed | m/s |
θ₀ | Launch angle above the horizontal | degrees |
v₀ₓ | Horizontal component of the launch velocity, constant throughout | m/s |
v₀ᵧ | Vertical component of the launch velocity | m/s |
x, y | Horizontal and vertical position | m |
x₀, y₀ | Launch position | m |
vᵧ | Vertical velocity at time t | m/s |
t | Time since launch | s |
g | Acceleration due to gravity | 9.80 m/s² |
R | Horizontal range on level ground | m |
h | Maximum height above the launch point | m |
When it applies
- Air resistance is small enough to ignore, which is the standard assumption in these problems.
- The range and time-of-flight shortcuts apply only when the launch and landing heights match. Otherwise go back to the component equations.
- After launch only gravity acts, so the horizontal velocity never changes and the vertical acceleration is a constant 9.8 m/s² downward.
Worked example
Problem. A ball is launched from level ground at 20 m/s, 30° above the horizontal. Find its time of flight, range and peak height.
- Split the launch velocity: v₀ₓ = 20 cos 30° = 17.3 m/s and v₀ᵧ = 20 sin 30° = 10.0 m/s.
- Time of flight on level ground: t = 2v₀ᵧ / g = 20.0 / 9.8 = 2.04 s.
- Range: R = v₀ₓt = (17.3)(2.04) = 35.3 m, which matches R = v₀² sin 2θ₀ / g = 400(0.866) / 9.8 = 35.3 m.
- Peak height: h = v₀ᵧ² / 2g = 100 / 19.6 = 5.1 m.
Answer. It stays in the air 2.0 s, lands about 35 m away, and peaks 5.1 m above the ground.
Common mistakes
- Putting gravity into the horizontal equation. The horizontal acceleration is zero.
- Concluding the ball is momentarily at rest at the top of the arc. The vertical velocity is zero there, but the horizontal component keeps going.
- Using the level-ground range formula when the projectile lands higher or lower than it started.
- Losing track of the time, which is the one quantity the horizontal and vertical equations share.
Related formulas
- Kinematics equations:
v = v₀ + at - Work-energy theorem:
W_net = ΔKE = ½mv² - ½mv₀² - Newton's second law:
ΣF = ma