Difficulty: Introductory to Intermediate | Prerequisites: 1D kinematics (position, velocity, acceleration), basic trigonometry (sine, cosine, tangent), vector components.
2D motion is where kinematics gets interesting, and where most students first struggle. The central idea is that motion in the x-direction and motion in the y-direction are independent of each other. This single principle unlocks both relative velocity problems (like swimming across a river with a current) and projectile motion (like a ball launched at an angle). If you are comfortable with 1D kinematics, the leap here is not a new set of equations, but learning to apply the old equations to each axis separately and then combine the results using vectors.
In 2D, treat x and y as separate 1D problems. For relative velocity, add velocity vectors (your velocity relative to the water plus the water's velocity relative to the ground). For projectile motion, the horizontal velocity stays constant while gravity only affects the vertical component. Use trig to split and recombine.
Relative velocity
The velocity of one object as observed from another moving object. Formally, v_AG = v_AW + v_WG, where A might be a swimmer, W the water, and G the ground. Think of it as: your motion as seen from the riverbank is your swimming motion plus the river's drift.
Vector components
The projections of a vector onto the x- and y-axes. Any 2D vector can be broken into v_x = v cos(θ) and v_y = v sin(θ). In simple terms, this means splitting diagonal motion into a horizontal part and a vertical part that you handle independently.
Projectile motion
The motion of an object launched into the air with some initial velocity, subject only to gravity (no air resistance). The horizontal and vertical motions are independent: gravity pulls only downward and has no effect on horizontal speed.
Maximum height
The highest point in a projectile's trajectory, where the vertical component of velocity is momentarily zero. The horizontal component is unaffected and stays at its initial value.
Range
The total horizontal distance a projectile covers from launch to landing. It depends on both the launch speed and the launch angle.
Launch angle (θ)
The angle between the initial velocity vector and the horizontal. It determines how the initial speed is split between horizontal and vertical components.
When you swim across a river, your velocity relative to the ground is the vector sum of your velocity relative to the water and the water's velocity relative to the ground.
If you aim straight across (perpendicular to the bank) but there is a current, you will arrive downstream of your target.
The angle θ between your actual path and the bank is found from the components: θ = tan⁻¹(v_water / v_swimmer).
Example from the exam: swimming at 1.5 m/s across a river with a 0.75 m/s current. The angle of drift is θ = tan⁻¹(0.75 / 1.5) = 27°.
To find how long it takes to cross, use only the component of velocity perpendicular to the banks: t = width / v_perpendicular. In the exam example, t = 25 m / 1.5 m/s = 17 s. The current affects where you land, not how long the crossing takes.
The core principle: horizontal and vertical motions are independent.
Horizontal (x): no acceleration (if air resistance is ignored), so v_x = v₀x = constant throughout the flight.
Vertical (y): constant downward acceleration due to gravity (g = 9.81 m/s²).
Initial velocity splits into components: v₀x = v₀ cos(θ) and v₀y = v₀ sin(θ).
The vertical velocity is zero: v_y = 0.
The horizontal velocity is unchanged: v_x = v₀x.
So the speed at the top equals the horizontal component of the initial velocity.
Example: v₀ = (25 m/s)x̂ + (16 m/s)ŷ. At the top, speed = v₀x = 25 m/s.
Maximum height: Δy = v₀y² / (2g) = (16)² / (2 × 9.81) = 13 m.
The horizontal component is still v₀x = 25 m/s.
The vertical component has the same magnitude as v₀y but is now negative (pointing downward): v_y = -v₀y = -16 m/s.
Total speed: v = √(v₀x² + v₀y²) = √(25² + 16²) = √(625 + 256) = √881 ≈ 30 m/s.
This equals the launch speed (for a projectile that lands at the same height it was launched from, the landing speed equals the launch speed).
Relative velocity (vector addition):
\vec{v}_{AG} = \vec{v}_{AW} + \vec{v}_{WG}Drift angle:
\theta = \tan^{-1}\left(\frac{v_{\text{water}}}{v_{\text{swimmer}}}\right)River crossing time (perpendicular swim):
t = \frac{w}{v_{\text{swimmer}}}Projectile components:
v_{0x} = v_0 \cos\theta \qquad v_{0y} = v_0 \sin\thetaMaximum height:
\Delta y_{\max} = \frac{v_{0y}^2}{2g}Speed at any point (from components):
v = \sqrt{v_x^2 + v_y^2}Vertical velocity at any time:
v_y = v_{0y} - g\,tRelative velocity is how pilots correct for crosswinds during landing, and how sailors account for ocean currents when plotting a course. Projectile motion governs everything from a basketball's arc to the trajectory of a water fountain's stream. Engineers designing irrigation sprinklers use these exact equations to calculate range and coverage.
Students often think the current affects how long it takes to cross the river. It does not (assuming you aim perpendicular to the bank). The current pushes you downstream, but crossing time depends only on the component of your velocity perpendicular to the banks.
Students often think the speed at the top of a projectile's arc is zero. It is not. The vertical component is zero, but the horizontal component is still v₀x. Speed is zero only if the projectile is thrown straight up.
Students often think that a projectile slows down in the horizontal direction as it rises. It does not. With no air resistance, horizontal velocity is constant throughout the flight.
Students often confuse speed and velocity at the landing point. Speed (a scalar) equals the launch speed for a level-ground projectile. Velocity (a vector) has the same magnitude but a different direction, because v_y has flipped sign.
⚠️ The river-crossing problem is a classic exam question. You will be expected to find the angle of drift (use tan⁻¹), the crossing time (use only the perpendicular velocity component), or both.
⚠️ For projectile motion, you must know the speed at the apex. The answer is always v₀x, the horizontal component. Examiners love offering 0 m/s as a distractor.
⚠️ Landing speed for a level-ground projectile equals the launch speed. The exam may ask you to compute it from components to verify you understand why.
⚠️ Maximum height uses v₀y² / (2g). Do not accidentally use the full launch speed v₀ in this formula.
True or false: a river current increases the time it takes to swim across if you aim perpendicular to the bank.
False. The current moves you downstream but does not change crossing time.
Fill in the blank: at the maximum height of a projectile, v_y = ______ and speed = ______.
v_y = 0; speed = v₀x (the horizontal component of the initial velocity).
True or false: a projectile launched at 30 m/s from level ground lands at 30 m/s.
True. The landing speed equals the launch speed when landing at the same height.
Fill in the blank: to find maximum height, use the formula Δy = ______ / ______.
v₀y² / (2g).
Q: You swim at 1.5 m/s in still water and aim straight across a 25 m wide river with a 0.75 m/s current. What angle does your actual path make with the direction you aimed?
A: θ = tan⁻¹(0.75 / 1.5) = tan⁻¹(0.5) = 27°.
Q: In the same river crossing, how long does it take to reach the other shore?
A: t = 25 m / 1.5 m/s = 17 s. The current does not affect crossing time when you aim perpendicular.
Q: A projectile is launched with v₀ = (25 m/s)x̂ + (16 m/s)ŷ. What is the maximum height above the launch point?
A: Δy = v₀y² / (2g) = (16)² / (2 × 9.81) = 256 / 19.62 = 13 m.
Q: What is the speed of the projectile at its maximum height?
A: At maximum height, v_y = 0, so speed = v₀x = 25 m/s.
Q: What is the speed of the projectile just before it lands (same height as launch)?
A: v = √(v₀x² + v₀y²) = √(25² + 16²) = √(625 + 256) = √881 ≈ 30 m/s. This equals the launch speed.
Relative velocity extends directly into reference frames and Galilean transformations if you continue in physics. Projectile motion is a special case of 2D kinematics with constant acceleration, which connects to circular motion (where acceleration is centripetal rather than gravitational). The independence of x and y components also appears in force analysis: Newton's second law is applied separately to each axis.
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