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Vectors and Projectile Motion

24 min Physics Math

In two dimensions, velocities and forces decompose into components using trigonometry.

Vector Components

$$v_x = v\cos\theta, \qquad v_y = v\sin\theta$$

$$|v| = \sqrt{v_x^2 + v_y^2}, \qquad \theta = \tan^{-1}(v_y/v_x)$$

Projectile Motion (no air resistance)

Horizontal: $x = v_0\cos\theta \cdot t$ (constant speed)

Vertical: $y = v_0\sin\theta \cdot t - \frac{1}{2}gt^2$ (free fall)

Range: $R = \frac{v_0^2 \sin 2\theta}{g}$

Example 1

A ball is launched at 40 m/s at 30° above horizontal. Find the range. ($g = 10$ m/s²)

$R = \frac{40^2 \sin 60°}{10} = \frac{1600 \times 0.866}{10} = 138.6$ m.

Example 2

Find the max height for the same projectile.

$v_y = v_0 \sin 30° = 20$ m/s.

$h = v_y^2/(2g) = 400/20 = 20$ m.

Example 3

A force of 50 N acts at 53° to the horizontal. Find horizontal and vertical components.

$F_x = 50\cos 53° \approx 50(0.6) = 30$ N.

$F_y = 50\sin 53° \approx 50(0.8) = 40$ N.

Practice Problems

1. Resolve a 100 m/s velocity at 60° into components.

2. Find the range of a projectile: $v_0 = 50$ m/s, $\theta = 45°$.

3. Find the magnitude of vector $(3, 4)$.

4. A ball is launched at 20 m/s at 30°. Find the max height.

5. Find the time of flight for $v_0 = 30$ m/s at 60°.

6. At what angle is range maximized?

Show Answer Key

1. $v_x = 50$ m/s, $v_y \approx 86.6$ m/s

2. $R = 2500/10 = 250$ m

3. $\sqrt{9+16} = 5$

4. $v_y = 10$, $h = 100/20 = 5$ m

5. $T = 2v_0 \sin 60°/g = 2(30)(0.866)/10 \approx 5.2$ s

6. 45°