Projectile Motion Formulas: Range, Height & Time

Last Updated: September 5, 2026

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Projectile motion formulas describe the two-dimensional motion of an object launched into the air when gravity is the only significant acceleration and air resistance is ignored.

For a projectile launched and landing at the same height, the three most important formulas are:

• Time of flight: (T = \frac{2u\sin\theta}{g})
• Maximum height: (H = \frac{u^2\sin^2\theta}{2g})
• Horizontal range: (R = \frac{u^2\sin(2\theta)}{g})

Before using these equations, resolve the initial velocity into horizontal and vertical components:

• (u_x = u\cos\theta)
• (u_y = u\sin\theta)

For a horizontal launch or a projectile landing at a different height, use the horizontal and vertical kinematic equations instead of the standard flat-ground range formula.

This guide explains the main projectile motion formulas, how the equations are derived and applied, horizontal and elevated launches, air resistance, worked examples, common mistakes, and practice questions.

Who This Projectile Motion Guide Is For

This projectile motion formulas guide is designed for high school and college physics students, AP Physics students, IGCSE and A-Level students, and anyone preparing for physics exams or solving introductory mechanics problems.

Use these formulas when you need to calculate:

• Projectile range on level ground
• Maximum height
• Time of flight
• Initial velocity components
• Horizontal distance
• Projectile velocity at a given time
• Trajectory and height at a specific distance
• Motion from a cliff, platform, or other elevated position

The guide is particularly useful for physics homework, exam preparation, classroom exercises, laboratory analysis, and introductory mechanics problem-solving.

Introduction to Projectile Motion

Projectile motion is a type of two-dimensional motion in which an object moves horizontally and vertically at the same time. In the ideal physics model, the projectile is acted on only by gravity after launch, so horizontal acceleration is zero while vertical acceleration is approximately (9.8\text{ m/s}^2) downward.

The key to solving projectile problems is to treat horizontal and vertical motion separately. The two components share the same time, but each follows its own motion equation.

If you are reviewing one-dimensional kinematics first, our SUVAT Equations Guide provides a useful foundation for applying kinematic equations to projectile problems.

What Is Projectile Motion?

Projectile motion occurs when an object is launched with an initial velocity at an angle to the horizontal and then follows a curved trajectory under gravity.

Examples include:

• A ball kicked into the air
• A basketball shot toward a hoop
• A stone thrown from a height
• A projectile launched from a platform
• A horizontally launched object falling from a cliff

Ignoring air resistance, the horizontal and vertical components behave independently.

Horizontal Motion

There is no horizontal acceleration in the ideal model:

[
a_x = 0
]

Therefore, horizontal velocity remains constant:

[
u_x = u\cos\theta
]

and horizontal displacement is:

[
x = u_x t
]

Vertical Motion

Gravity produces constant downward acceleration:

[
a_y = -g
]

Using (g = 9.8\text{ m/s}^2), the vertical velocity is:

[
v_y = u\sin\theta – gt
]

and the vertical displacement is:

[
y = u\sin\theta , t – \frac{1}{2}gt^2
]

This separation of horizontal and vertical motion is the foundation of projectile motion calculations.

Projectile Motion Formula Sheet

The following equations cover the most common ideal projectile motion problems.

Quantity	Formula	Main condition
Horizontal velocity	(u_x = u\cos\theta)	No horizontal acceleration
Vertical velocity	(u_y = u\sin\theta)	Initial vertical component
Horizontal displacement	(x = u_x t)	(a_x=0)
Vertical displacement	(y = u_y t-\frac12gt^2)	Constant gravity
Vertical velocity at time (t)	(v_y=u_y-gt)	Constant gravity
Time of flight	(T=\frac{2u\sin\theta}{g})	Same launch and landing height
Maximum height	(H=\frac{u^2\sin^2\theta}{2g})	Height above launch point
Horizontal range	(R=\frac{u^2\sin(2\theta)}{g})	Same launch and landing height

Resolving Initial Velocity Into Components

When a projectile is launched at speed (u) and angle (\theta), the initial velocity must be separated into horizontal and vertical components.

Horizontal Component

[
u_x=u\cos\theta
]

Vertical Component

[
u_y=u\sin\theta
]

This step is essential because gravity affects only the vertical component in the ideal model.

For example, if an object is launched at (20\text{ m/s}) at (30^\circ):

[
u_x=20\cos30^\circ\approx17.32\text{ m/s}
]

[
u_y=20\sin30^\circ=10.0\text{ m/s}
]

Once these components are known, the projectile can be analyzed using separate horizontal and vertical equations.

The Three Main Projectile Motion Formulas

1. Time of Flight Formula

For a projectile launched from ground level and landing at the same height:

[
T=\frac{2u\sin\theta}{g}
]

Where:

• (T) = total time of flight in seconds
• (u) = initial speed in m/s
• (\theta) = launch angle
• (g) = gravitational acceleration

This equation comes from the vertical motion. At the end of the flight, the projectile has returned to its original vertical position.

The formula does not apply directly when the projectile lands at a different height.

2. Maximum Height Formula

At maximum height, the vertical velocity is zero:

[
v_y=0
]

The maximum height above the launch point is:

[
H=\frac{u^2\sin^2\theta}{2g}
]

An equivalent form using the initial vertical velocity is:

[
H=\frac{u_y^2}{2g}
]

This formula is useful when the vertical component of the initial velocity is already known.

3. Horizontal Range Formula

For a projectile that launches and lands at the same vertical level:

[
R=\frac{u^2\sin(2\theta)}{g}
]

Where R is the horizontal distance from launch to landing.

This formula depends on three main factors:

• Initial speed
• Launch angle
• Gravitational acceleration

The equation should not be used unchanged for an elevated launch or a landing point at a different height.

Why Is the Maximum Range at 45 Degrees?

For equal launch and landing heights:

[
R=\frac{u^2\sin(2\theta)}{g}
]

The maximum value of sine is 1. Therefore:

[
\sin(2\theta)=1
]

which occurs when:

[
2\theta=90^\circ
]

so:

[
\theta=45^\circ
]

Therefore, 45° gives the maximum horizontal range for a given initial speed when launch and landing heights are equal, and air resistance is ignored.

This condition is important because the 45° result does not apply universally to every projectile situation.

Projectile Motion Kinematic Equations

Projectile motion can also be solved directly with kinematic equations.

Horizontal Equations

Because:

[
a_x=0
]

the horizontal displacement is:

[
x=u_xt
]

Since:

[
u_x=u\cos\theta
]

we get:

[
x=u\cos\theta , t
]

The horizontal velocity remains:

[
v_x=u_x=u\cos\theta
]

Vertical Equations

The vertical acceleration is:

[
a_y=-g
]

The vertical velocity after time (t) is:

[
v_y=u\sin\theta-gt
]

The vertical displacement is:

[
y=u\sin\theta,t-\frac12gt^2
]

Another useful equation is:

[
v_y^2=(u\sin\theta)^2-2gy
]

These equations are particularly useful when the projectile does not launch and land at the same height.

Projectile Trajectory Equation

The trajectory describes the height of a projectile as a function of horizontal distance.

For an ideal projectile launched from the origin:

[
y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}
]

The equation has an (x^2) term, which produces the familiar parabolic path.

The trajectory equation is useful when a problem asks for the projectile’s height at a particular horizontal distance rather than its total range.

Symmetry in Projectile Motion

Ideal projectile motion has several useful symmetry properties when the launch and landing heights are equal.

Time Symmetry

The time required to reach maximum height is half the total flight time:

[
t_{\text{peak}}=\frac{T}{2}
]

Velocity Symmetry

At equal heights during ascent and descent, the vertical velocity has the same magnitude but opposite direction.

Complementary Angles

For the same initial speed and equal launch and landing heights, complementary launch angles such as (30^\circ) and (60^\circ) produce the same horizontal range.

They do not produce the same maximum height or time of flight.

Effect of Launch Angle on Range

For a fixed initial speed and equal launch and landing heights:

Launch Angle	(\sin(2\theta))	Relative Range
15°	0.50	50% of maximum
30°	0.87	87% of maximum
45°	1.00	Maximum
60°	0.87	87% of maximum
75°	0.50	50% of maximum
90°	0.00	Zero horizontal range

The important point is that 30° and 60° produce the same range under these ideal conditions, while 45° produces the maximum range.

Projectile Launched Horizontally From a Height

A horizontal launch is a special case because:

[
\theta=0^\circ
]

Therefore:

[
u_y=0
]

Suppose an object is launched horizontally from height (h) with speed (u).

Step 1: Find Time to Reach the Ground

The vertical motion starts from rest:

[
h=\frac12gt^2
]

Therefore:

[
t=\sqrt{\frac{2h}{g}}
]

Step 2: Find Horizontal Range

Horizontal speed remains constant:

[
R=ut
]

Substituting the time:

[
R=u\sqrt{\frac{2h}{g}}
]

Step 3: Find Vertical Velocity at Impact

[
v_y=gt
]

The final speed can then be found from the horizontal and vertical components:

[
v=\sqrt{u_x^2+v_y^2}
]

This approach is more appropriate than using the standard flat-ground range formula.

Projectile Landing at a Different Height

If a projectile lands above or below its launch point, the standard range formula

[
R=\frac{u^2\sin(2\theta)}{g}
]

cannot be used directly.

Instead, use the vertical equation:

[
y=u_y t-\frac12gt^2
]

where:

[
u_y=u\sin\theta
]

Solve for time first. Then calculate horizontal distance using:

[
x=u_x t
]

where:

[
u_x=u\cos\theta
]

This method works for cliffs, towers, elevated platforms, and other non-level launch and landing points.

Air Resistance and Real Projectile Motion

The standard projectile motion formulas assume negligible air resistance.

In real-world motion, aerodynamic drag opposes the projectile’s movement and changes its trajectory.

Air resistance can:

• Reduce horizontal range
• Reduce maximum height
• Change the flight time
• Make the trajectory asymmetric
• Change the launch angle that produces maximum range

Therefore, the familiar 45° maximum-range result applies to the ideal equal-height, no-air-resistance model.

For classroom and exam problems, air resistance is commonly ignored unless the question specifically includes it.

Students can also use an interactive projectile simulation to compare ideal trajectories with different physical conditions.

Worked Example 1: Find Range, Height, and Time

A ball is launched at:

[
u=20\text{ m/s}
]

at:

[
\theta=30^\circ
]

Assume:

[
g=10\text{ m/s}^2
]

and that the ball lands at the same height.

Step 1: Find the Velocity Components

Horizontal component:

[
u_x=20\cos30^\circ
]

[
u_x=17.32\text{ m/s}
]

Vertical component:

[
u_y=20\sin30^\circ
]

[
u_y=10.0\text{ m/s}
]

Step 2: Calculate Time of Flight

[
T=\frac{2u\sin\theta}{g}
]

[
T=\frac{2(20)(0.5)}{10}
]

[
\boxed{T=2.0\text{ s}}
]

Step 3: Calculate Maximum Height

[
H=\frac{u_y^2}{2g}
]

[
H=\frac{10^2}{2(10)}
]

[
\boxed{H=5.0\text{ m}}
]

Step 4: Calculate Range

[
R=u_xT
]

[
R=17.32(2.0)
]

[
\boxed{R=34.64\text{ m}}
]

Therefore:

• Time of flight = 2.0 s
• Maximum height = 5.0 m
• Horizontal range = 34.64 m

Worked Example 2: Horizontal Launch From a Cliff

A ball is launched horizontally at:

[
u=10\text{ m/s}
]

from a height of:

[
h=45\text{ m}
]

Use:

[
g=10\text{ m/s}^2
]

Step 1: Calculate Time

[
t=\sqrt{\frac{2h}{g}}
]

[
t=\sqrt{\frac{2(45)}{10}}
]

[
t=\sqrt9
]

[
\boxed{t=3.0\text{ s}}
]

Step 2: Calculate Horizontal Distance

[
x=ut
]

[
x=10(3)
]

[
\boxed{x=30.0\text{ m}}
]

The ball reaches the ground after 3 seconds and travels 30 meters horizontally.

Worked Example 3: Maximum Range at 45 Degrees

A golf ball is launched at:

[
u=30\text{ m/s}
]

at:

[
\theta=45^\circ
]

Use:

[
g=10\text{ m/s}^2
]

Maximum Height

[
H=\frac{u^2\sin^245^\circ}{2g}
]

Since:

[
\sin45^\circ\approx0.707
]

we get:

[
H=\frac{900(0.5)}{20}
]

[
\boxed{H=22.5\text{ m}}
]

Horizontal Range

[
R=\frac{u^2\sin90^\circ}{g}
]

[
R=\frac{900(1)}{10}
]

[
\boxed{R=90.0\text{ m}}
]

The 45° launch produces the maximum range for this ideal equal-height situation.

How to Solve Projectile Motion Problems

A reliable method is:

Step 1: Identify the Given Values

Write down:

• Initial speed (u)
• Launch angle (\theta)
• Initial height
• Final height
• Gravitational acceleration (g)

Step 2: Resolve the Initial Velocity

Calculate:

[
u_x=u\cos\theta
]

[
u_y=u\sin\theta
]

Step 3: Decide Which Motion Controls the Unknown

Use vertical motion for:

• Time
• Maximum height
• Vertical velocity

Use horizontal motion for:

• Horizontal distance
• Range

Step 4: Check the Launch and Landing Heights

If they are equal, the standard time and range formulas may apply.

If they are different, use the full kinematic equations.

Step 5: Check Units

Use consistent SI units:

• Speed: m/s
• Distance: m
• Time: s
• Acceleration: m/s²
• Angle: degrees when using standard trigonometric calculations

Common Projectile Motion Mistakes

Using the Range Formula for an Elevated Launch

The equation:

[
R=\frac{u^2\sin(2\theta)}{g}
]

requires equal launch and landing heights.

Forgetting Velocity Components

Do not use the full initial speed in a vertical equation when the problem requires the vertical component.

Use:

[
u_y=u\sin\theta
]

and:

[
u_x=u\cos\theta
]

Mixing Horizontal and Vertical Equations

Horizontal and vertical motion are analyzed separately. The common variable connecting them is time.

Using the Wrong Sign for Gravity

If upward is positive, then:

[
a_y=-g
]

A consistent sign convention prevents many calculation errors.

Ignoring Units

Always check that velocity, distance, time, and acceleration use compatible units before calculating.

15 Projectile Motion Practice MCQs

Q1. What is the horizontal acceleration of a projectile when air resistance is ignored?

A) (g) downward
B) (g) upward
C) Zero
D) It depends on launch speed

Q2. Which launch angle gives maximum range on flat ground?

A) 30°
B) 45°
C) 60°
D) 90°

Q3. At maximum height, the vertical component of velocity is:

A) Maximum
B) Equal to the horizontal component
C) Zero
D) Negative

Q4. A ball is thrown horizontally at 20 m/s and takes 3 seconds to land. What is its horizontal range?

A) 20 m
B) 60 m
C) 30 m
D) 9.8 m

Q5. A projectile is launched at 30 m/s at 45°. What is its initial horizontal velocity component?

A) 30 m/s
B) 15 m/s
C) 21.2 m/s
D) 25.9 m/s

Q6. Launch angles of 30° and 60° with the same initial speed and equal launch and landing heights result in:

A) Same range, same height
B) Different range, same height
C) Same range, different height
D) Different range, different height

Q7. A projectile’s path is parabolic in the ideal model because:

A) Both horizontal and vertical motion are accelerated
B) Horizontal motion is uniform while vertical motion is uniformly accelerated
C) Only vertical motion is uniform
D) Air resistance creates the curve

Q8. Which equation gives the time of flight for a projectile launched from and landing on the same level?

A) (T=u/g)
B) (T=2u\sin\theta/g)
C) (T=u\sin\theta/g)
D) (T=2u/(g\sin\theta))

Q9. A ball is thrown at 40 m/s at 30°. What is its maximum height? Use (g=10\text{ m/s}^2).

A) 10 m
B) 20 m
C) 40 m
D) 80 m

Q10. What happens to the horizontal component of velocity during ideal projectile motion?

A) It increases
B) It decreases
C) It remains constant
D) It becomes zero at maximum height

Q11. A stone is dropped from a bridge. Its initial vertical velocity is:

A) (u\sin\theta)
B) (u\cos\theta)
C) (g)
D) Zero

Q12. Projectile motion can be analyzed by treating horizontal and vertical motion:

A) Together using one equation
B) Independently
C) Only when air resistance is included
D) Only for angles below 45°

Q13. A cannonball is fired at 50 m/s at 37°. Find its range using (g=10\text{ m/s}^2), (\sin37^\circ=0.6), and (\cos37^\circ=0.8).

A) 120 m
B) 200 m
C) 240 m
D) 300 m

Q14. In ideal projectile motion, speed is minimum at:

A) Launch
B) Maximum height
C) Landing
D) It remains equal throughout

Q15. A ball is launched horizontally from a 20 m tower. Use (g=10\text{ m/s}^2). What is the time to hit the ground?

A) 1 s
B) 2 s
C) 4 s
D) 10 s

Answers

1-C, 2-B, 3-C, 4-B, 5-C, 6-C, 7-B, 8-B, 9-B, 10-C, 11-D, 12-B, 13-C, 14-B, 15-B

For Q9:

[
u_y=40\sin30^\circ=20\text{ m/s}
]

[
H=\frac{20^2}{2(10)}=20\text{ m}
]

Therefore, the correct answer is B.

Frequently Asked Questions About Projectile Motion Formulas

What are the three main projectile motion formulas?

For a projectile launched and landing at the same height:

[
T=\frac{2u\sin\theta}{g}
]

is the time of flight,

[
H=\frac{u^2\sin^2\theta}{2g}
]

is the maximum height, and

[
R=\frac{u^2\sin(2\theta)}{g}
]

is the horizontal range.

What is the formula for projectile motion range?

For equal launch and landing heights:

[
R=\frac{u^2\sin(2\theta)}{g}
]

The formula should not be used directly when the projectile lands at a different height.

What is the formula for maximum height in projectile motion?

The maximum height above the launch point is:

[
H=\frac{u^2\sin^2\theta}{2g}
]

or, using the initial vertical velocity:

[
H=\frac{u_y^2}{2g}
]

What is the time of flight formula?

For equal launch and landing heights:

[
T=\frac{2u\sin\theta}{g}
]

For an elevated launch or a different landing height, solve the vertical kinematic equation instead.

What is the projectile velocity formula?

The initial velocity can be resolved into:

[
u_x=u\cos\theta
]

and:

[
u_y=u\sin\theta
]

At time (t), the vertical velocity is:

[
v_y=u\sin\theta-gt
]

while the horizontal velocity remains constant in the ideal model.

Does mass affect projectile motion?

In the ideal vacuum model, mass does not change the projectile’s acceleration, time of flight, maximum height, or range when the initial conditions are the same. Gravity produces the same acceleration for objects regardless of mass.

What is the speed at maximum height?

At maximum height:

[
v_y=0
]

The horizontal component remains:

[
v_x=u\cos\theta
]

Therefore, for the ideal model, the speed at maximum height is:

[
v=u\cos\theta
]

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

For equal launch and landing heights, the range depends on:

[
\sin(2\theta)
]

Since:

[
\sin(60^\circ)=\sin(120^\circ)
]

30° and 60° produce the same range for the same initial speed. Their maximum heights and flight times are different.

Can energy conservation be used for projectile motion?

Yes. When air resistance is ignored, mechanical energy can be used to relate kinetic and gravitational potential energy.

The relationship is:

[
KE+PE=\text{constant}
]

This can be useful for determining speed at different heights.

Final Projectile Motion Formula Checklist

Before solving a problem, ask:

1. Is air resistance being ignored?
2. Are the launch and landing heights equal?
3. What is the initial speed?
4. What is the launch angle?
5. Have I resolved the velocity into (u_x) and (u_y)?
6. Do I need time, height, range, velocity, or trajectory?
7. Should I use the standard formula or full kinematic equations?
8. Are my units consistent?
9. Is my sign convention consistent?
10. Does my final answer make physical sense?

The most important formulas to remember are:

[
u_x=u\cos\theta
]

[
u_y=u\sin\theta
]

[
T=\frac{2u\sin\theta}{g}
]

[
H=\frac{u^2\sin^2\theta}{2g}
]

[
R=\frac{u^2\sin(2\theta)}{g}
]

and:

[
y=u\sin\theta,t-\frac12gt^2
]

For equal-height launches and landings, the first three formulas provide a fast way to calculate the main characteristics of ideal projectile motion. For horizontal or elevated launches, use the component-based kinematic equations.