Kinematics Calculator (Uniform Motion, Acceleration, Free Fall, Projectile)

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projectile motion calculator range maximum height

Projectile motion calculator: range, maximum height and flight time

Oblique projectile motion combines two independent movements: a uniform one horizontally and a uniformly accelerated one vertically. Give the speed and angle, or any two of range, maximum height and flight time, and the calculator draws the path and the vectors.

How range and height depend on the angle

Split the initial velocity: vₓ = v₀·cosθ (constant) and v_y0 = v₀·sinθ (which gravity reduces by g every second). The maximum height is reached when v_y becomes zero, at t = v_y0/g, with H = h₀ + v_y0²/(2g). Launched from the ground, the time of flight is 2·v_y0/g and the range is R = v₀²·sin2θ/g. With v₀ = 20 m/s, θ = 30° and g = 10 m/s², vₓ = 17.32 m/s, v_y0 = 10 m/s, the highest point is 5 m, the flight lasts 2 s and the range is 34.64 m.
On the ground, the range is greatest at 45° and equals v₀²/g, which is 40 m for 20 m/s and g = 10 m/s². Complementary angles give the same range, because sin2θ = sin(180° - 2θ): 30° and 60° both reach 34.64 m, but the 60° path climbs higher and takes longer. For a 30 m range at 20 m/s, the angles are 24.3° and 65.7°. The calculator shows both and dashes the second trajectory.
If the launch starts from a height h₀, the time of flight becomes T = [v_y0 + √(v_y0² + 2·g·h₀)]/g and the angle of maximum range falls below 45°, with tan θ = v₀/√(v₀² + 2·g·h₀) and maximum range (v₀/g)·√(v₀² + 2·g·h₀). With 20 m/s, h₀ = 10 m and g = 10 m/s², the best angle is 39.2° and the maximum range is 48.99 m. A horizontal launch is the case θ = 0: from an 80 m cliff at 15 m/s, the fall takes 4 s and the range is 60 m.
The model is the ideal one, with no air resistance. In reality air shortens the range, makes the descent steeper than the climb and pulls the best angle below 45°. The formulas above are the ones from high school and introductory physics, and work well for heavy balls and short distances.

Six launches worked out

Launch at 45°

Input: v₀ = 20 m/s, θ = 45°, g = 10 m/s²
Expected output: R = 40 m, H = 10 m, T ≈ 2,83 s

Maximum ground range: v₀²/g.

Launch at 30°

Input: v₀ = 20 m/s, θ = 30°, g = 10 m/s²
Expected output: R ≈ 34,64 m, H = 5 m, T = 2 s

The 60° angle gives the same range with H = 15 m.

Two angles for 30 m

Input: v₀ = 20 m/s, R = 30 m, g = 10 m/s²
Expected output: θ ≈ 24,3° e θ ≈ 65,7°

sin2θ = 10·30/400 = 0.75.

Speed required

Input: θ = 45°, R = 40 m, g = 10 m/s²
Expected output: v₀ = 20 m/s

v₀ = √(g·R/sin2θ).

Best angle from 10 m

Input: v₀ = 20 m/s, h₀ = 10 m, g = 10 m/s²
Expected output: θ ≈ 39,2°, R máx ≈ 48,99 m

Below 45° because of the starting height.

Horizontal launch from the cliff

Input: v₀ = 15 m/s, h₀ = 80 m, g = 10 m/s²
Expected output: T = 4 s, R = 60 m

It lands at √(15² + 40²) ≈ 42.7 m/s.

Full tool FAQ

What is the difference between uniform motion and constant acceleration?

In uniform motion the velocity does not change (zero acceleration) and position grows along a straight line on the s-t graph: S = S₀ + v·t. With constant acceleration the acceleration is non-zero, velocity changes linearly, v = v₀ + a·t, and the s-t graph is a parabola: S = S₀ + v₀·t + a·t²/2. Uniform motion is the special case with a = 0. A car at a steady 72 km/h covers 100 m in 5 s; one starting from rest with a = 2 m/s² covers 25 m in the same 5 s.

How do I convert km/h to m/s?

When should I use Torricelli's equation?

Which g should I use: 9.8, 10 or 9.80665?

How long does a free fall take and how fast does the object arrive?

How do I find the range and maximum height of an oblique launch?

Why do two different angles give the same range?

Why is average speed not the mean of the speeds?

What do two solutions or the negative time warning mean?

Does the calculator include air resistance?

Frequently asked questions

Why does 45° give the longest range?

Because the range is proportional to sin2θ, which reaches its maximum of 1 when 2θ = 90°. At smaller angles the projectile is fast horizontally but lands soon; at larger ones it stays long in the air but travels little sideways. This holds on the ground and without air.

Is the landing speed equal to the launch speed?

Only if the ground is at the same height as the launch: then the landing angle also equals the launch angle. Launching from a height h₀, the final speed is √(v₀² + 2·g·h₀), larger than the initial one.

Can I find when the projectile passes a given height?

Yes, in the height query field. With 20 m/s at 30° and g = 10 m/s², the height of 4 m is crossed twice, going up and coming down; and 5 m is the highest point, crossed once, at t = 1 s. Heights above H are never reached.

Next steps

How range and height depend on the angle

Split the initial velocity: vₓ = v₀·cosθ (constant) and v_y0 = v₀·sinθ (which gravity reduces by g every second). The maximum height is reached when v_y becomes zero, at t = v_y0/g, with H = h₀ + v_y0²/(2g). Launched from the ground, the time of flight is 2·v_y0/g and the range is R = v₀²·sin2θ/g. With v₀ = 20 m/s, θ = 30° and g = 10 m/s², vₓ = 17.32 m/s, v_y0 = 10 m/s, the highest point is 5 m, the flight lasts 2 s and the range is 34.64 m.

On the ground, the range is greatest at 45° and equals v₀²/g, which is 40 m for 20 m/s and g = 10 m/s². Complementary angles give the same range, because sin2θ = sin(180° - 2θ): 30° and 60° both reach 34.64 m, but the 60° path climbs higher and takes longer. For a 30 m range at 20 m/s, the angles are 24.3° and 65.7°. The calculator shows both and dashes the second trajectory.

If the launch starts from a height h₀, the time of flight becomes T = [v_y0 + √(v_y0² + 2·g·h₀)]/g and the angle of maximum range falls below 45°, with tan θ = v₀/√(v₀² + 2·g·h₀) and maximum range (v₀/g)·√(v₀² + 2·g·h₀). With 20 m/s, h₀ = 10 m and g = 10 m/s², the best angle is 39.2° and the maximum range is 48.99 m. A horizontal launch is the case θ = 0: from an 80 m cliff at 15 m/s, the fall takes 4 s and the range is 60 m.

The model is the ideal one, with no air resistance. In reality air shortens the range, makes the descent steeper than the climb and pulls the best angle below 45°. The formulas above are the ones from high school and introductory physics, and work well for heavy balls and short distances.

Six launches worked out

Launch at 45°

Input: v₀ = 20 m/s, θ = 45°, g = 10 m/s²
Expected output: R = 40 m, H = 10 m, T ≈ 2,83 s

Maximum ground range: v₀²/g.

Launch at 30°

Input: v₀ = 20 m/s, θ = 30°, g = 10 m/s²
Expected output: R ≈ 34,64 m, H = 5 m, T = 2 s

The 60° angle gives the same range with H = 15 m.

Two angles for 30 m

Input: v₀ = 20 m/s, R = 30 m, g = 10 m/s²
Expected output: θ ≈ 24,3° e θ ≈ 65,7°

sin2θ = 10·30/400 = 0.75.

Speed required

Input: θ = 45°, R = 40 m, g = 10 m/s²
Expected output: v₀ = 20 m/s

v₀ = √(g·R/sin2θ).

Best angle from 10 m

Input: v₀ = 20 m/s, h₀ = 10 m, g = 10 m/s²
Expected output: θ ≈ 39,2°, R máx ≈ 48,99 m

Below 45° because of the starting height.

Horizontal launch from the cliff

Input: v₀ = 15 m/s, h₀ = 80 m, g = 10 m/s²
Expected output: T = 4 s, R = 60 m

It lands at √(15² + 40²) ≈ 42.7 m/s.

Full tool FAQ

What is the difference between uniform motion and constant acceleration?

How do I convert km/h to m/s?

When should I use Torricelli's equation?

Which g should I use: 9.8, 10 or 9.80665?

How long does a free fall take and how fast does the object arrive?

How do I find the range and maximum height of an oblique launch?

Why do two different angles give the same range?

Why is average speed not the mean of the speeds?

What do two solutions or the negative time warning mean?

Does the calculator include air resistance?

Frequently asked questions

Continue in the tool

More pages for this tool

Projectile motion calculator: range, maximum height and flight time

How range and height depend on the angle

Six launches worked out

Launch at 45°

Launch at 30°

Two angles for 30 m

Speed required

Best angle from 10 m

Horizontal launch from the cliff

Full tool FAQ

Frequently asked questions

Why does 45° give the longest range?

Is the landing speed equal to the launch speed?

Can I find when the projectile passes a given height?

Next steps

Problem data

Positions (optional)

Result

Unit conversion (to SI)

Step by step

Graphs

Position vs time

Velocity vs time

Acceleration vs time

Continue in the tool

More pages for this tool

Projectile motion calculator: range, maximum height and flight time

How range and height depend on the angle

Six launches worked out

Launch at 45°

Launch at 30°

Two angles for 30 m

Speed required

Best angle from 10 m

Horizontal launch from the cliff

Full tool FAQ

Frequently asked questions

Why does 45° give the longest range?

Is the landing speed equal to the launch speed?

Can I find when the projectile passes a given height?

Next steps

Problem data

Positions (optional)

Result

Unit conversion (to SI)

Step by step

Graphs

Position vs time

Velocity vs time

Acceleration vs time