Projectile Motion

By the end of this lesson, you will be able to analyze two-dimensional projectile motion by decomposing it into independent horizontal and vertical components.

The Key Insight: Independence of Motion

Horizontal and vertical motions are completely independent.

• Horizontal: constant velocity (no air resistance → no horizontal acceleration)
• Vertical: constant acceleration due to gravity ($a_y = -g = -9.8$ m/s²)

This independence means you solve two separate 1D problems simultaneously, connected only by time.

Setting Up Projectile Problems

Launched at speed $v_0$ at angle $\theta$ above horizontal:

$$v_{0x} = v_0\cos\theta \qquad v_{0y} = v_0\sin\theta$$

Horizontal:

$$x = v_{0x}t$$

Vertical:

$$y = v_{0y}t - \frac{1}{2}gt^2 \qquad v_y = v_{0y} - gt$$

Example 1 — Launched at an Angle

A soccer ball is kicked at 25 m/s at 35° above horizontal. Find: (a) time of flight, (b) range (horizontal distance), (c) max height.

$v_{0x} = 25\cos 35° = 20.5$ m/s; $v_{0y} = 25\sin 35° = 14.3$ m/s

(a) Time of flight (lands at same height → $y = 0$):

$0 = 14.3t - 4.9t^2 \implies t(14.3 - 4.9t) = 0 \implies t = 2.92$ s

(b) Range: $x = v_{0x}t = 20.5 \times 2.92 = 59.9$ m

(c) Max height (at $t = v_{0y}/g = 14.3/9.8 = 1.46$ s):

$y = 14.3(1.46) - 4.9(1.46)^2 = 20.9 - 10.4 = 10.5$ m

Example 2 — Horizontal Launch

A ball is launched horizontally at 15 m/s from a cliff 40 m high. Find (a) time to hit ground, (b) horizontal range.

$v_{0x} = 15$ m/s; $v_{0y} = 0$ (horizontal launch)

(a) $-40 = 0 - \frac{1}{2}(9.8)t^2 \implies t = \sqrt{80/9.8} = 2.86$ s

(b) $x = 15 \times 2.86 = 42.9$ m

The Range Equation

For projectile landing at the same height:

$$R = \frac{v_0^2\sin(2\theta)}{g}$$

Maximum range at $\theta = 45°$ (since $\sin 90° = 1$).

Complementary angles give the same range: 30° and 60° give identical horizontal range (but different heights and times).

Summary

Two independent components: horizontal (constant $v_x$); vertical ($a_y = -g$).

Time connects both components: use vertical to find time; use time for horizontal.

$v_{0y} = 0$; vertical is pure free fall.