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Topic 4 of 20

Two-Dimensional Kinematics

Extending kinematics to two dimensions uses vectors and the key insight that motion in perpendicular directions is completely independent. Projectile motion — the most important 2D kinematics application — follows from combining constant horizontal velocity with constant downward free-fall acceleration.

Key Concepts

Independence of Motion

Horizontal and vertical motions are independent. Horizontal: constant velocity (no air resistance). Vertical: constant acceleration

g

downward. Time is the only shared variable linking the two.

Projectile Motion

A launched object follows a parabolic path. Initial velocity

v_0

at angle

\theta

gives

v_{0x} = v_0\cos\theta

and

v_{0y} = v_0\sin\theta

. Horizontal speed is constant; vertical speed changes due to gravity.

Time of Flight

Total time in the air is determined entirely by the vertical motion. For a projectile returning to its launch height:

T = 2v_0\sin\theta/g

Range

The horizontal distance traveled over the full flight:

R = v_0^2\sin 2\theta / g

. Maximum range occurs at launch angle

\theta = 45°

(on level ground).

Maximum Height

The peak height above the launch point, reached when

v_y = 0

H = v_0^2\sin^2\theta / (2g)

Relative Velocity

The velocity of object A as seen by observer C:

\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}

. Velocities add vectorially — always specify the reference frame.

Key Equations

Horizontal position

x = v_{0x}\,t = v_0\cos\theta\;t

Constant horizontal velocity; no horizontal acceleration (ignoring air resistance).

Vertical position

y = v_{0y}\,t - \tfrac{1}{2}gt^2 = v_0\sin\theta\;t - \tfrac{1}{2}gt^2

Free-fall equation for the vertical coordinate.

Vertical velocity

v_y = v_0\sin\theta - gt

Vertical velocity decreases linearly; zero at the peak.

Range (level ground)

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

Horizontal distance when the projectile lands at the same height it was launched.

Maximum height

H = \frac{v_0^2\sin^2\!\theta}{2g}

Peak height above the launch point.

Worked Example

Projectile Launched at an Angle

Problem

A ball is kicked at $v_0 = 20$ m/s at $\theta = 30°$ above horizontal. Find the range and maximum height (take $g = 10$ m/s²).

Solution

Find initial components:

v_{0x} = 20\cos30° = 17.3 \text{ m/s}, \quad v_{0y} = 20\sin30° = 10 \text{ m/s}

Maximum height (at $v_y = 0$ ):

H = \frac{v_{0y}^2}{2g} = \frac{(10)^2}{2(10)} = 5 \text{ m}

Range:

R = \frac{v_0^2\sin 2\theta}{g} = \frac{(20)^2\sin 60°}{10} = \frac{400 \times 0.866}{10} \approx 34.6 \text{ m}

Answer Maximum height H = 5 m; range R ≈ 34.6 m.

Practice

Exercises

7 problems

1 of 7

A projectile is launched at $v_0 = 20 \text{ m/s}$ at $\theta = 30°$ above horizontal. What is the initial horizontal velocity component? (Use $\cos30° \approx 0.866$ )

Your answer

m/s

2 of 7

The same projectile ( $v_0 = 20 \text{ m/s}$ , $\theta = 30°$ ). What is the initial vertical velocity component? (Use $\sin30° = 0.5$ )

Your answer

m/s

3 of 7

Using the same projectile ( $v_{0y} = 10 \text{ m/s}$ , $g = 10 \text{ m/s}^2$ ). What is the maximum height above the launch point?

Your answer

4 of 7

Using the same projectile ( $v_{0y} = 10 \text{ m/s}$ , $g = 10 \text{ m/s}^2$ ). What is the total time of flight?

Your answer

5 of 7

Using the same projectile ( $v_{0x} = 17.3 \text{ m/s}$ , $T = 2 \text{ s}$ ). What is the horizontal range?

Your answer

6 of 7

A ball is kicked at $15 \text{ m/s}$ at $45°$ ( $g = 10 \text{ m/s}^2$ ). What is the horizontal range?

Your answer

7 of 7

A projectile is launched horizontally at $25 \text{ m/s}$ from a cliff $20 \text{ m}$ high ( $g = 10 \text{ m/s}^2$ ). How long does it take to hit the ground?

Your answer

Key Takeaways

In 2D kinematics, always resolve the initial velocity into horizontal and vertical components first.
Horizontal: no acceleration (constant $v_x$ ). Vertical: constant $g$ downward.
Time of flight depends only on the vertical motion — solve the vertical equation for $t$ first when needed.
Range is maximized at 45° and complementary angles (e.g., 30° and 60°) give the same range.
Relative velocity: $\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$ — always add vectorially.