Two-Dimensional Motion: Quick Notes Two-Dimensional Motion: Independence of x and y Base equations and relationships in 2D/3D are unchanged; x and y components act independently. Treat as two separate 1D problems (one for x, one for y) and combine results vectorially when needed. Decomposing Motion into Components Break any 2D problem into its x and y parts. Solve each part with its own initial velocity and acceleration, then recombine. Final velocity components: v < e m > x , v < / e m > y v<em>x,\; v</em>y v < e m > x , v < / e m > y Magnitude: v = v < e m > x 2 + v < / e m > y 2 v = \sqrt{v<em>x^2 + v</em>y^2} v = v < e m > x 2 + v < / e m > y 2 Direction: θ = tan − 1 ( v < e m > y v < / e m > x ) \theta = \tan^{-1}\left(\frac{v<em>y}{v</em>x}\right) θ = tan − 1 ( v < / e m > x v < e m > y ) Projectile Motion Basics (gravity only) a < e m > x = 0 , v < / e m > x ( t ) = v 0 x a<em>x = 0,\quad v</em>x(t) = v_{0x} a < e m > x = 0 , v < / e m > x ( t ) = v 0 x a < e m > y = − g , v < / e m > y ( t ) = v 0 y − g t a<em>y = -g,\quad v</em>y(t) = v_{0y} - g t a < e m > y = − g , v < / e m > y ( t ) = v 0 y − g t y ( t ) = v 0 y t − 1 2 g t 2 y(t) = v_{0y} t - \tfrac{1}{2} g t^2 y ( t ) = v 0 y t − 2 1 g t 2 g ≈ 9.8 m / s 2 g \approx 9.8\ \mathrm{m/s^2} g ≈ 9.8 m/ s 2 Example Problem: Separate x and y (7 s) Given: v < e m > 0 x = 22 m / s , a < / e m > x = 24 m / s 2 ; v < e m > 0 y = 14 m / s , a < / e m > y = 12 m / s 2 ; t = 7 s v<em>{0x}=22\ \mathrm{m/s},\; a</em>x=24\ \mathrm{m/s^2};\quad v<em>{0y}=14\ \mathrm{m/s},\; a</em>y=12\ \mathrm{m/s^2};\; t=7\ \mathrm{s} v < e m > 0 x = 22 m/s , a < / e m > x = 24 m/ s 2 ; v < e m > 0 y = 14 m/s , a < / e m > y = 12 m/ s 2 ; t = 7 s Displacements: x = v < e m > 0 x t + 1 2 a < / e m > x t 2 x = v<em>{0x} t + \tfrac{1}{2} a</em>x t^2 x = v < e m > 0 x t + 2 1 a < / e m > x t 2 y = v < e m > 0 y t + 1 2 a < / e m > y t 2 y = v<em>{0y} t + \tfrac{1}{2} a</em>y t^2 y = v < e m > 0 y t + 2 1 a < / e m > y t 2 Final components:v < e m > x = v < / e m > 0 x + a x t = 190 m / s v<em>x = v</em>{0x} + a_x t = 190\ \mathrm{m/s} v < e m > x = v < / e m > 0 x + a x t = 190 m/s v < e m > y = v < / e m > 0 y + a y t = 98 m / s v<em>y = v</em>{0y} + a_y t = 98\ \mathrm{m/s} v < e m > y = v < / e m > 0 y + a y t = 98 m/s Final speed: v = v < e m > x 2 + v < / e m > y 2 ≈ 214 m / s v = \sqrt{v<em>x^2 + v</em>y^2} \approx 214\ \mathrm{m/s} v = v < e m > x 2 + v < / e m > y 2 ≈ 214 m/s Direction: θ = tan − 1 ( v < e m > y v < / e m > x ) ≈ 27.5 ∘ \theta = \tan^{-1}\left(\frac{v<em>y}{v</em>x}\right) \approx 27.5^{\circ} θ = tan − 1 ( v < / e m > x v < e m > y ) ≈ 27. 5 ∘ Final Velocity Vector (components) Velocity components: v < e m > x = 190 m / s , v < / e m > y = 98 m / s v<em>x = 190\ \mathrm{m/s},\quad v</em>y = 98\ \mathrm{m/s} v < e m > x = 190 m/s , v < / e m > y = 98 m/s Vector form can give magnitude and angle as above Time of Flight and Range: Horizontal drop example Scenario: plane speed v 0 x = 115 m / s v_{0x}=115\ \mathrm{m/s} v 0 x = 115 m/s H = 1050 m H=1050\ \mathrm{m} H = 1050 m Time in air: t = 2 H g ≈ 14.6 s t = \sqrt{\dfrac{2H}{g}} \approx 14.6\ \mathrm{s} t = g 2 H ≈ 14.6 s Final velocities: v < e m > x = 115 m / s , v < / e m > y = − g t ≈ − 143 m / s v<em>x = 115\ \mathrm{m/s},\quad v</em>y = -g t \approx -143\ \mathrm{m/s} v < e m > x = 115 m/s , v < / e m > y = − g t ≈ − 143 m/s Final speed: v = 115 2 + 143 2 ≈ 184 m / s v = \sqrt{115^2 + 143^2} \approx 184\ \mathrm{m/s} v = 11 5 2 + 14 3 2 ≈ 184 m/s Flight angle: θ = tan − 1 ( v < e m > y v < / e m > x ) ≈ − 51.2 ∘ \theta = \tan^{-1}\left(\dfrac{v<em>y}{v</em>x}\right) \approx -51.2^{\circ} θ = tan − 1 ( v < / e m > x v < e m > y ) ≈ − 51. 2 ∘ Horizontal range: R = v x t ≈ 1.68 × 10 3 m R = v_x t \approx 1.68 \times 10^3\ \mathrm{m} R = v x t ≈ 1.68 × 1 0 3 m Strategy and Tips for 2D Motion Problems Draw a diagram; set positive directions clearly. Treat x and y as separate problems; verify at least three kinetic values per component. Break any initial velocity given as magnitude + angle into components: v < e m > 0 x = v < / e m > 0 cos θ , v < e m > 0 y = v < / e m > 0 sin θ v<em>{0x} = v</em>0\cos\theta,\; v<em>{0y} = v</em>0\sin\theta v < e m > 0 x = v < / e m > 0 cos θ , v < e m > 0 y = v < / e m > 0 sin θ If motion occurs in segments (varying accelerations), solve per segment and chain results. Combine x and y results to obtain the final vector (magnitude and direction). Note: air resistance is neglected in these baseline problems; real motion may differ. Quick takeaway Any 2D motion can be analyzed by independently solving x(t) and y(t) using their own initial conditions and accelerations, then recombining to get the overall motion.