Core idea

Projectile motion is two independent one-dimensional motions sharing the same time t. Ignore air resistance unless stated otherwise.

Choose a positive direction and keep signs consistent. With right/up positive:

aₓ = 0 and aᵧ = −g = −9.8 m/s²

Components

For launch speed v₀ at angle θ above horizontal:

v₀ₓ = v₀ cos θ v₀ᵧ = v₀ sin θ

For a horizontal launch, v₀ᵧ = 0.

Equations

Horizontal

aₓ = 0 vₓ = v₀ₓ Δx = v₀ₓt

Vertical

Use constant-acceleration kinematics:

vᵧ = v₀ᵧ − gt Δy = v₀ᵧt − ½gt² vᵧ² = v₀ᵧ² − 2gΔy

Where the equations come from

Start with the constant-acceleration equations:

v = v₀ + at Δs = v₀t + ½at² v² = v₀² + 2aΔs

Apply them separately to each direction.

Horizontal derivation

Since aₓ = 0:

vₓ = v₀ₓ + 0t = v₀ₓ

Then:

Δx = v₀ₓt + ½(0)t² = v₀ₓt

Vertical derivation

Since aᵧ = −g:

vᵧ = v₀ᵧ + (−g)t = v₀ᵧ − gt

Δy = v₀ᵧt + ½(−g)t² = v₀ᵧt − ½gt²

vᵧ² = v₀ᵧ² + 2(−g)Δy = v₀ᵧ² − 2gΔy

Component derivation

Resolve the launch-velocity vector into a right triangle:

cos θ = v₀ₓ/v₀ → v₀ₓ = v₀ cos θ sin θ = v₀ᵧ/v₀ → v₀ᵧ = v₀ sin θ

Useful shortcuts derived from the equations

Time to the highest point

At the top, vᵧ = 0:

0 = v₀ᵧ − gt tᵤₚ = v₀ᵧ/g

Maximum height

Use vᵧ² = v₀ᵧ² − 2gΔy and set vᵧ = 0:

0 = v₀ᵧ² − 2gΔyₘₐₓ Δyₘₐₓ = v₀ᵧ²/(2g)

Total flight time: same launch and landing height

Set Δy = 0 in the vertical-position equation:

0 = v₀ᵧt − ½gt² 0 = t(v₀ᵧ − ½gt)

Ignoring the launch instant t = 0:

t_total = 2v₀ᵧ/g

Range: same launch and landing height

Use Δx = v₀ₓt and substitute t_total = 2v₀ᵧ/g:

R = (v₀ₓ)(2v₀ᵧ/g)

Substitute the components:

R = (v₀ cos θ)(2v₀ sin θ)/g

Using 2 sin θ cos θ = sin 2θ:

R = v₀² sin(2θ)/g

These same-height shortcuts do not apply when the launch and landing heights differ.

Fast solve method

1. Draw the trajectory and choose positive directions.
2. Split the initial velocity into x- and y-components.
3. Solve the vertical motion for the shared time t or the needed vᵧ.
4. Use that same t in the horizontal equation.
5. Check units, signs, and whether the answer is reasonable.

Special cases

Horizontal launch from height h

With downward positive for the vertical motion:

t = √(2h/g) Δx = vₓt

The horizontal speed does not affect fall time.

Highest point

vᵧ = 0, but vₓ is still nonzero and gravity is still −g.

For launch and landing at the same height:

tᵤₚ = t_down vᵧf = −v₀ᵧ

Maximum height above the launch point:

Δyₘₐₓ = v₀ᵧ²/(2g)

Graphs

• x vs. t: straight line; slope vₓ.
• y vs. t: parabola; curvature comes from gravity.
• vᵧ vs. t: straight line; slope −g.

Common traps

• t is shared by both directions; do not use separate times.
• Gravity changes vertical motion only, not vₓ.
• At the top, vᵧ = 0, not the total velocity.
• Δy means final height minus initial height; it can be negative.
• Do not use same-height shortcuts when launch and landing heights differ.