Projectile Motion Calculator — range, height & time of flight

How it works

Projectile motion treats the horizontal and vertical directions separately: the horizontal velocity is constant while gravity accelerates the object downward. For a launch over level ground at speed v₀ and angle θ, OpenStax University Physics gives three results:

• Horizontal range — R = v₀² × sin(2θ) ÷ g
• Maximum height — H = v₀² × sin²θ ÷ (2g)
• Time of flight — T = 2 × v₀ × sinθ ÷ g

where g is the acceleration due to gravity (9.81 m/s² near Earth's surface). Speed is in metres per second and the angle in degrees, giving range and height in metres and time in seconds.

Those three formulas assume the projectile lands back at its launch height. This calculator also accepts an initial height h₀ — launching from a cliff, table or platform. The projectile then lands on the ground below, so it stays airborne longer and travels farther. Solving the vertical motion h₀ + v₀ × sinθ × t − ½ × g × t² = 0 for the landing time gives t = (v₀ × sinθ + √((v₀ × sinθ)² + 2 × g × h₀)) ÷ g; the range is R = v₀ × cosθ × t, and the peak sits h₀ above the ground at H = h₀ + v₀² × sin²θ ÷ (2g). With h₀ = 0 these reduce to the level-ground formulas above.

Two consequences fall straight out of the range formula. Because sin(2θ) peaks at 2θ = 90°, the range is greatest at a 45° launch angle. And because sin(2θ) is the same for θ and 90° − θ, a projectile travels the same horizontal distance at complementary angles — 15° and 75°, or 30° and 60° — one on a low fast arc, the other high and slow. The model ignores air resistance, so real projectiles fall a little short of these ideal figures.

Worked example

Launched at 20 m/s and 45° with g = 9.81 m/s²:

• Range: R = 20² × sin(90°) ÷ 9.81 = 400 ÷ 9.81 ≈ 40.77 m
• Maximum height: H = 400 × sin²(45°) ÷ (2 × 9.81) = 200 ÷ 19.62 ≈ 10.19 m
• Time of flight: T = 2 × 20 × sin(45°) ÷ 9.81 ≈ 2.88 s

Frequently asked questions

What is the range of a projectile?

The range is the horizontal distance a projectile covers before returning to its launch height. Over level ground it is R = v₀² × sin(2θ) ÷ g, where v₀ is the launch speed, θ the angle and g gravity.

What launch angle gives the maximum range?

45°. The range depends on sin(2θ), which is largest when 2θ = 90°, i.e. θ = 45° — assuming launch and landing are at the same height and air resistance is ignored.

Why do 30° and 60° give the same range?

Because sin(2θ) is identical for an angle and its complement: sin(60°) = sin(120°). The lower angle produces a flat, fast trajectory and the higher one a tall, slow arc, but both land the same distance away.

Does this account for air resistance?

No. These formulas assume ideal projectile motion — gravity only, no drag. Air resistance shortens the real range and lowers the peak, especially for light or fast objects.

How long is the projectile in the air?

The time of flight is T = 2 × v₀ × sinθ ÷ g. Only the vertical part of the launch velocity matters, so a steeper launch stays airborne longer for the same speed.

How high does the projectile go?

Its maximum height is H = v₀² × sin²θ ÷ (2g), reached at the top of the arc when the vertical velocity is momentarily zero. A steeper launch angle gives a higher peak; a 90° (straight-up) launch reaches the greatest height of all, v₀² ÷ (2g). Launched from an initial height h₀, the peak sits that much higher: H = h₀ + v₀² × sin²θ ÷ (2g).

Can I launch the projectile from a height?

Yes. Set the **initial height** h₀ to launch from a cliff, roof or platform; the projectile lands on the ground below, so it stays in the air longer and travels farther than a level-ground launch at the same speed and angle. Leave h₀ = 0 when the launch and landing are at the same height.