Projectile Motion Formula

Projectile motion is two-dimensional motion under gravity alone, where horizontal velocity is constant and vertical motion is uniformly accelerated —.

The Formula

x=v0cos⁡θ⋅t;y=v0sin⁡θ⋅t−12gt2x = v_0 \cos\theta \cdot t \quad ; \quad y = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2

When to use: A thrown ball follows a curved path—horizontal motion is steady, vertical is accelerated.

Quick Example

A ball thrown at 45°45° travels farthest; at 90°90° it goes straight up and down.

Notation

v0v_0 is the initial speed in m/s, θ\theta is the launch angle, g≈9.81g \approx 9.81 m/s² is gravitational acceleration, xx is horizontal position, yy is vertical position, and RR is the range.

What This Formula Means

Two-dimensional motion under gravity alone, where horizontal velocity is constant and vertical motion is uniformly accelerated — producing a parabolic path.

A thrown ball follows a curved path—horizontal motion is steady, vertical is accelerated.

Formal View

With launch angle θ\theta and initial speed v0v_0, the position is x(t)=v0cos⁡θ⋅tx(t) = v_0 \cos\theta \cdot t and y(t)=v0sin⁡θ⋅t−12gt2y(t) = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2. The range on level ground is R=v02sin⁡2θgR = \frac{v_0^2 \sin 2\theta}{g}, maximised at θ=45°\theta = 45°.

Worked Examples

Example 1

medium
A ball is launched horizontally at 15 m/s15 \text{ m/s} from a cliff 45 m45 \text{ m} high. How far from the base of the cliff does it land? Use g=10 m/s2g = 10 \text{ m/s}^2.

Answer

x=45 mx = 45 \text{ m}

First step

1
Use vertical motion to find the time to fall: t=2hg=2×4510=9=3 st = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{10}} = \sqrt{9} = 3 \text{ s}

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Example 2

hard
A projectile is launched at 40 m/s40 \text{ m/s} at 30°30° above the horizontal. What is the maximum height and horizontal range? Use g=10 m/s2g = 10 \text{ m/s}^2.

Example 3

medium
A projectile is launched at 30 m/s30 \text{ m/s} at 30°30° above the horizontal (g=10g = 10, sin⁡30°=0.5\sin 30° = 0.5). Find the maximum height above launch.

Common Mistakes

  • Applying acceleration in the horizontal direction — there is no horizontal acceleration in ideal projectile motion (air resistance is neglected), so horizontal velocity is constant. - Fix this by naming the system, checking "Am I describing motion over time with position, distance, direction, speed, velocity, or acceleration clearly separated?", and attaching units or direction to the final statement.
  • Using the full initial speed instead of its components — you must resolve v0v_0 into v0x=v0cos⁡θv_{0x} = v_0\cos\theta and v0y=v0sin⁡θv_{0y} = v_0\sin\theta before applying kinematic equations. - Fix this by naming the system, checking "Am I describing motion over time with position, distance, direction, speed, velocity, or acceleration clearly separated?", and attaching units or direction to the final statement.
  • Forgetting that at the highest point the vertical velocity is zero but the horizontal velocity is not — the projectile does not stop at the top, it only stops rising. - Fix this by naming the system, checking "Am I describing motion over time with position, distance, direction, speed, velocity, or acceleration clearly separated?", and attaching units or direction to the final statement.
  • Using projectile motion from a keyword alone - Signal words like position, speed, velocity only point to a possible model; the system must match too.

Why This Formula Matters

Projectile Motion helps students describe motion precisely instead of relying on everyday words like fast or slow. It prepares them to interpret graphs, choose equations, and connect motion to forces and energy.

Frequently Asked Questions

What is the Projectile Motion formula?

Two-dimensional motion under gravity alone, where horizontal velocity is constant and vertical motion is uniformly accelerated — producing a parabolic path.

How do you use the Projectile Motion formula?

A thrown ball follows a curved path—horizontal motion is steady, vertical is accelerated.

What do the symbols mean in the Projectile Motion formula?

v0v_0 is the initial speed in m/s, θ\theta is the launch angle, g≈9.81g \approx 9.81 m/s² is gravitational acceleration, xx is horizontal position, yy is vertical position, and RR is the range.

Why is the Projectile Motion formula important in Physics?

Projectile Motion helps students describe motion precisely instead of relying on everyday words like fast or slow. It prepares them to interpret graphs, choose equations, and connect motion to forces and energy.

What do students get wrong about Projectile Motion?

Students often know a formula related to projectile motion but skip the recognition step: Am I describing motion over time with position, distance, direction, speed, velocity, or acceleration clearly separated? That leads to a correct-looking substitution attached to the wrong physical model.

What should I learn before the Projectile Motion formula?

Before studying the Projectile Motion formula, you should understand: free fall, velocity, vectors.

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