Projectile Motion Formula

The Formula

x = 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° travels farthest; at 90° it goes straight up and down.

Notation

v_0 is the initial speed in m/s, \theta is the launch angle, g \approx 9.81 m/s² is gravitational acceleration, x is horizontal position, y is vertical position, and R 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 \theta and initial speed v_0, the position is x(t) = v_0 \cos\theta \cdot t and y(t) = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2. The range on level ground is R = \frac{v_0^2 \sin 2\theta}{g}, maximised at \theta = 45°.

Worked Examples

Example 1

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

Solution

  1. 1
    Use vertical motion to find the time to fall: t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{10}} = \sqrt{9} = 3 \text{ s}
  2. 2
    Horizontal speed stays constant because there is no horizontal acceleration.
  3. 3
    Horizontal distance: x = v_x \cdot t = 15 \times 3 = 45 \text{ m}

Answer

x = 45 \text{ m}
In projectile motion, horizontal and vertical motions are independent. The horizontal velocity remains constant (no air resistance), while the vertical motion is free fall.

Example 2

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

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.
  • Using the full initial speed instead of its components — you must resolve v_0 into v_{0x} = v_0\cos\theta and v_{0y} = v_0\sin\theta before applying kinematic equations.
  • 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.

Why This Formula Matters

Projectile motion models the trajectory of thrown balls, kicked footballs, launched rockets, and fired bullets. It is central to sports science, military ballistics, space mission planning, and the design of fountains and firework displays.

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?

v_0 is the initial speed in m/s, \theta is the launch angle, g \approx 9.81 m/s² is gravitational acceleration, x is horizontal position, y is vertical position, and R is the range.

Why is the Projectile Motion formula important in Physics?

Projectile motion models the trajectory of thrown balls, kicked footballs, launched rockets, and fired bullets. It is central to sports science, military ballistics, space mission planning, and the design of fountains and firework displays.

What do students get wrong about Projectile Motion?

The horizontal velocity stays constant (no horizontal acceleration).

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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