MechanicalMechanics

Projectile Motion

Separate horizontal and vertical motion to understand the whole trajectory

Difficulty

Beginner

Read time

6 min

Review status

Needs review

Concept sketch

Projectile Motion

Illustrative model, not experimental data

Overview

A projectile follows a curved path because gravity changes its vertical velocity while its horizontal velocity stays constant in the simplest model. Treating the two directions separately explains flight time, range, and peak height without treating the curve as a new force.

How to read it

Locate the launch and landing heights.
Read horizontal distance and vertical height separately.
At the peak, check vertical velocity rather than total speed.

When to use it

Estimating an ideal launch trajectory.
Comparing launch angles at a fixed speed.

Choose the model before the equations

Model a particle moving near Earth's surface with constant downward gravitational acceleration. Ignore air resistance, spin, and propulsion. These assumptions are often reasonable for a classroom trajectory, but not for a long-range or high-drag flight.

Resolve the launch velocity

Measure the launch angle θ above horizontal. The horizontal component is u cos θ and the vertical component is u sin θ. With the launch point at x = 0 and height y₀, time ties the two directions together.

x (t) = u cos θ t y (t) = y_{0} + u sin θ t - \frac{1}{2} g t^{2}

The peak is not a complete stop

At the highest point the vertical velocity is zero, but horizontal velocity remains u cos θ. Acceleration is still downward with magnitude g; it does not vanish at the top.

Know when the range shortcut applies

The familiar range expression requires landing at the launch height. At different landing heights, first solve the vertical equation for the positive impact time, then use horizontal speed times time.

R = \frac{u ^{2} sin ( 2 θ )}{g} (y_{landing} = y_{0})

Worked example

A ball launches at 20 m/s, 30° above level ground, and lands at the same height. Ignore drag and take g = 9.81 m/s².

1.Resolve the vertical speed and find the total flight time.
$u_{y} = 20 sin 3 0^{\circ} = 10 m/s t_{f} = \frac{2 ( 10 )}{9.81} = 2.039 s$
2.Multiply the horizontal speed by that time.
$R = (20 cos 3 0^{\circ}) (2.039) = 35.31 m$

The ideal flight lasts about 2.04 s and covers 35.3 m.

Key takeaways

Horizontal and vertical motion share time.
Gravity acts throughout the flight.
A 45° launch maximizes equal-height range only in the no-drag model.

Common mistakes

Using the equal-height range formula for an elevated launch.
Making gravity act horizontally.
Setting total velocity to zero at the peak.

Related formulas

Range of a Projectile Newton's Second Law of Motion

Related concepts

Free-Body Diagrams Numerical Kinematics

Source

Aiman Amri — projectile_motion.ipynb

Adapted from the notebook model under MIT. Explanations, diagrams, and worked examples were created for FormuLab.

Adapted from Aiman Amri, Mechanical Engineering Calculations.

MIT license and copyright notice

Rewritten as a concept lesson with explicit assumptions, original diagrams, and a new worked example. External images, executable code, and notebook outputs are not included.

Projectile Motion

Projectile Motion

How to read it

When to use it

Choose the model before the equations

Resolve the launch velocity

The peak is not a complete stop

Know when the range shortcut applies

Worked example

Product

Company

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