Projectile Motion: Equations, Range and Max Height

Projectile motion is motion under gravity alone after launch. For launch speed ( v_0 ) at angle ( theta ): ( x = v_0costheta,t ) and ( y = h_0 + v_0sintheta,t – frac12gt^2 ). On level ground the time of flight is ( T = frac{2v_0sintheta}{g} ), the range is ( R = frac{v_0^2sin 2theta}{g} ) and the maximum height is ( H = frac{v_0^2sin^2theta}{2g} ). The path is a parabola, and ( 45^circ ) gives the longest range.

Projectile motion is the motion of an object launched into the air and then moving under gravity alone: a thrown ball, a kicked football, water from a fountain. Ignoring air resistance, the path is always a parabola, and every question about it (how long, how far, how high) can be answered with a handful of equations.

The key idea is that horizontal and vertical motion are independent. Horizontally nothing pushes or pulls, so velocity stays constant. Vertically gravity gives a constant downward acceleration ( g approx 9.81 ) m/s².

Step 1: split the velocity into components

A projectile launched at speed ( v_0 ) and angle ( theta ) above the horizontal has

$$v_x = v_0costheta, qquad v_y = v_0sintheta$$

This is right-triangle trigonometry: the launch velocity is the hypotenuse (see trigonometric functions).

Step 2: write position as a function of time

Starting at height ( h_0 ):

$$x(t) = v_0costhetacdot t$$

$$y(t) = h_0 + v_0sinthetacdot t - tfrac12gt^2$$

The vertical velocity is ( v_y(t) = v_0sintheta – gt ).

Where these come from

They are integrals. The acceleration is ( a_x = 0 ) and ( a_y = -g ). Integrating once gives the velocities, using the initial velocity as the constant. Integrating again gives the positions, using the starting point as the constant. It’s the same process as the one-dimensional kinematic equations, applied in each direction.

Step 3: the key results

For a launch and landing at the same height (( h_0 = 0 )):

Quantity	Formula	How it’s found
Time to peak	( t_{text{peak}} = frac{v_0sintheta}{g} )	set ( v_y = 0 )
Maximum height	( H = frac{v_0^2sin^2theta}{2g} )	( y(t_{text{peak}}) )
Time of flight	( T = frac{2v_0sintheta}{g} )	set ( y = 0 ), ( t neq 0 )
Range	( R = frac{v_0^2sin 2theta}{g} )	( x(T) )

The range formula uses the double-angle identity ( 2sinthetacostheta = sin 2theta ) (see trig identities).

The trajectory is a parabola

Solve ( x = v_0costhetacdot t ) for t and substitute into ( y ):

$$y = h_0 + xtantheta - frac{g}{2v_0^2cos^2theta}x^2$$

That’s a quadratic in x with a negative leading coefficient: a downward parabola. Its vertex is the maximum height.

Example 1: a kicked ball

A football is kicked at 25 m/s at ( 30^circ ) from the ground. Find the time of flight, range and maximum height.

Components. ( v_x = 25cos 30^circ approx 21.65 ) m/s, ( v_y = 25sin 30^circ = 12.5 ) m/s.

Time of flight. ( T = frac{2(12.5)}{9.81} approx 2.548 ) s.

Range. ( R = 21.65 times 2.548 approx 55.17 ) m. (Check: ( frac{625sin 60^circ}{9.81} approx 55.17 ).)

Maximum height. ( H = frac{12.5^2}{2(9.81)} approx 7.96 ) m, at ( t approx 1.274 ) s.

Example 2: launched horizontally from a cliff

A stone is thrown horizontally at 15 m/s from a cliff 20 m high. How far from the base does it land?

With ( theta = 0 ), there’s no initial vertical velocity, so the fall time comes from the vertical motion alone:

$$20 = tfrac12(9.81)t^2 quadRightarrowquad t = sqrt{frac{40}{9.81}} approx 2.019 text{ s}$$

Horizontal distance: ( 15 times 2.019 approx 30.29 ) m. The stone takes exactly as long to land as one dropped from rest. The horizontal throw doesn’t change the fall time.

Example 3: launch from above the ground

A ball is thrown at 20 m/s at ( 60^circ ) from a height of 1.5 m. When does it land?

Here ( v_y = 20sin 60^circ approx 17.32 ) m/s. Set ( y = 0 ):

$$1.5 + 17.32t - 4.905t^2 = 0$$

The quadratic formula gives ( t approx 3.616 ) s (the other root is negative). Then ( R = 10 times 3.616 approx 36.16 ) m, since ( v_x = 20cos 60^circ = 10 ) m/s.

Example 4: the best angle

On level ground, which angle gives the longest range? Treat ( R(theta) = frac{v_0^2sin 2theta}{g} ) as a function and maximize it (optimization):

$$R'(theta) = frac{2v_0^2cos 2theta}{g} = 0 quadRightarrowquad 2theta = 90^circ quadRightarrowquad theta = 45^circ$$

Angles equally above and below ( 45^circ ) give the same range: ( 30^circ ) and ( 60^circ ) land at the same spot, because ( sin 60^circ = sin 120^circ ). The ( 60^circ ) shot goes higher and stays up longer.

Projectile motion calculator

Enter launch speed, angle and height to get everything at once. Try ( 30^circ ) and ( 60^circ ) at the same speed to see the equal ranges.

The full-page projectile motion calculator has more worked examples.

Velocity at any moment

The speed at time t is ( sqrt{v_x^2 + v_y(t)^2} ), and the direction is ( arctanfrac{v_y}{v_x} ) (see inverse trig functions). At the top, ( v_y = 0 ), so the speed is at its minimum, ( v_x ). On level ground the projectile lands with the same speed it was launched with, angled downward at the launch angle.

Common mistakes

• Using the range formula from a height. ( R = frac{v_0^2sin 2theta}{g} ) only applies when launch and landing heights match. Otherwise solve ( y(t) = 0 ).
• Thinking the velocity is zero at the top. Only the vertical part is zero; the horizontal part never changes.
• Calculator in radian mode. ( cos 30 ) in radians is about 0.154, not 0.866.
• Applying gravity horizontally. ( a_x = 0 ) without air resistance.
• Forgetting air resistance exists. Real balls fall short of these predictions, and the best angle is lower than ( 45^circ ).

Practice problems

Use ( g = 9.81 ) m/s² and ignore air resistance.

1. A ball is launched at 30 m/s at ( 45^circ ) on level ground. Find its range.
2. For the ball in problem 1, find the maximum height.
3. A package is dropped from a plane flying horizontally at 50 m/s, 80 m above the ground. How long does it fall?
4. In problem 3, how far ahead of the drop point does it land?
5. A ball is thrown at 10 m/s at ( 90^circ ) (straight up). How long until it returns?

Answers: (1) about 91.74 m; (2) about 22.94 m; (3) about 4.039 s; (4) about 201.9 m; (5) about 2.039 s.

FAQ

What is projectile motion?

The motion of an object moving freely under gravity after launch. Horizontal velocity stays constant and vertical motion has constant acceleration, so the path is a parabola.

What are the projectile motion equations?

( x = v_0costheta,t ) and ( y = h_0 + v_0sintheta,t – frac12gt^2 ). Everything else follows from these.

Why is the horizontal velocity constant?

No horizontal force acts on the projectile (ignoring air), so by Newton’s first law its horizontal velocity doesn’t change.

Does a heavier object travel farther?

Not without air resistance. Mass doesn’t appear in any of the equations.

What is the angle for maximum height?

( 90^circ ), straight up. Maximum height grows with ( sin^2theta ), but the range is then zero.

Further reading

• 4.3 Projectile Motion (OpenStax University Physics Volume 1) — derivations of time of flight, range and trajectory with examples.
• Projectile motion (Wikipedia) — equations, the parabolic trajectory and models with air resistance.