What is the formula for range of a projectile?

For level-ground launch and zero air resistance, range R = v₀² sin(2θ) / g. Maximum range is at θ = 45° on level ground.

How do I find the maximum height of a projectile?

At the apex the vertical velocity is zero, so h = v₀² sin²(θ) / (2g) above the launch point. Add the launch height to get height above the ground.

Why are horizontal and vertical motion independent?

Gravity acts vertically and there are no other forces (we ignore air resistance), so horizontal velocity is unchanged while vertical velocity changes by g each second. The two axes share only the time of flight.

What value of g should I use?

On Earth's surface, g = 9.8 m/s². Use 9.81 if the question gives that, or 9.80 for older NESA papers. The calculator defaults to 9.8.

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NSW · HSCModule 5

Projectile motion calculator

Enter the launch speed, angle and height. Get the range, maximum height, time of flight, and a sketch of the parabolic trajectory. Built for HSC Physics Module 5.

Inputs

Initial speed v₀ (m/s)Launch angle θ (degrees)Above the horizontalLaunch height (m)0 for level groundGravitational acceleration g (m/s²)Earth surface ≈ 9.8

Result

Range

62.81m

Max height

13.18m

Time of flight

3.280s

vₓ (constant)

19.15m/s

v_y at launch

16.07m/s

Time to apex

1.640s

0x = 62.8 m

How this calculator works

A projectile launched at speed v₀ and angle θ above the horizontal has horizontal component v₀ cos θ and vertical component v₀ sin θ. Horizontal velocity is constant; vertical velocity changes by g each second.

The calculator computes:

Time of flight by solving the vertical equation y = h + v_y t − ½ g t² for y = 0.
Range from horizontal velocity × time of flight.
Maximum height from v_y² = 0 at the apex, giving h_max = h + v_y² / (2g).

Want the full worked-example explanation? Read our projectile motion dot point answer.

Common questions

What is the formula for range of a projectile?: For level-ground launch and zero air resistance, range R = v₀² sin(2θ) / g. Maximum range is at θ = 45° on level ground.
How do I find the maximum height of a projectile?: At the apex the vertical velocity is zero, so h = v₀² sin²(θ) / (2g) above the launch point. Add the launch height to get height above the ground.
Why are horizontal and vertical motion independent?: Gravity acts vertically and there are no other forces (we ignore air resistance), so horizontal velocity is unchanged while vertical velocity changes by g each second. The two axes share only the time of flight.
What value of g should I use?: On Earth's surface, g = 9.8 m/s². Use 9.81 if the question gives that, or 9.80 for older NESA papers. The calculator defaults to 9.8.