Kinematics 2D

Two-dimensional motion requires two coordinates to describe position at every instant. We use the Cartesian system with $\hat{i}$ (east/right) and $\hat{j}$ (north/up) as unit vectors.

Position vector: $\vec{r} = x\hat{i} + y\hat{j}$

Displacement: $\Delta\vec{r} = \vec{r}_2 - \vec{r}_1 = \Delta x\hat{i} + \Delta y\hat{j}$

The key principle of 2D kinematics: x and y motions are completely independent. This is the foundation of projectile motion analysis. An x-component of force affects only x-motion; a y-component of force affects only y-motion.

A projectile is any object launched into the air with an initial velocity and subsequently acted upon only by gravity (air resistance neglected). The horizontal and vertical motions are treated separately.

Let the projectile be launched from the origin at angle θ with speed u. At launch: $u_x = u\cos\theta$, $u_y = u\sin\theta$.

Time of flight (returns to launch height, y = 0):

Maximum height (at $v_y = 0$, i.e., $t = u\sin\theta/g$):

Horizontal range ($x$ when $y = 0$):

Range is maximum at $\theta = 45°$, giving $R_{max} = u^2/g$.

Complementary angles (θ and 90°−θ) give the same range since $\sin2\theta = \sin(180°-2\theta)$.

The trajectory is the path of the projectile in the x-y plane. We derive it by eliminating time from the parametric equations.

From horizontal: $t = \frac{x}{u\cos\theta}$. Substitute into vertical:

This can be rewritten using $\sec^2\theta = 1 + \tan^2\theta$:

This is a quadratic in x with a negative leading coefficient — it defines a downward-opening parabola.

A horizontal projectile is launched with $u_y = 0$ (purely horizontal). It is the special case $\theta = 0$.

Tower problems involve a projectile launched from a cliff or building of height H. Key ideas:

• For a dropped or horizontal throw, vertical fall time is $t = \sqrt{2H/g}$.
• For an upward throw from the roof, set displacement = −H (below launch point) and solve the quadratic for t.
• Two bodies projected horizontally from the same height at the same time hit the ground simultaneously regardless of initial speed.

When a projectile is launched on a slope of angle α, the standard reference frame (horizontal-vertical) is still the most reliable approach. Alternatively, a tilted frame aligned with the slope can simplify some calculations.

Setup: Incline at angle α to horizontal. Projectile launched at angle β above horizontal (so β − α above the slope).

Key formulas for projection up the incline:

Angle for maximum range up the slope: $\beta = 45° + \alpha/2$ (above horizontal), or $45° - \alpha/2$ above the incline surface.

Maximum range on incline: $R_{max} = \frac{u^2}{g(1+\sin\alpha)}$

For projection down the incline, replace α with −α in the formulas above. The angle for maximum range down the slope is $\beta = 45° - \alpha/2$ above horizontal.

Relative velocity: The velocity of object A as seen by observer B is $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$. This is a vector subtraction.

River crossing: A boat of speed $v_b$ (in still water) crosses a river of width d with current $v_r$.

Rain-Man problems: Rain falls with velocity $\vec{v}_{rain}$; man walks with velocity $\vec{v}_{man}$. The apparent velocity of rain to the man is $\vec{v}_{rain/man} = \vec{v}_{rain} - \vec{v}_{man}$. The man should tilt the umbrella in the direction of $\vec{v}_{rain/man}$ (the direction the rain appears to come from).