Projectile motion — horizontal - vertical independence, full derivation

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WHY does independence hold? (first principles)

WHAT this means: $a_x=0$ ⇒ horizontal velocity never changes. $a_y=-g$ ⇒ vertical velocity changes exactly like a dropped object. Neither equation contains the other variable — they are decoupled.

HOW we exploit it: solve each axis separately with 1-D kinematics, then combine using the shared $t$.

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Full derivation from scratch

Launch from origin with speed $u$ at angle $\theta$ above horizontal. Split the initial velocity: $u_x = u\cos\theta, \qquad u_y = u\sin\theta$

Step 1 — Integrate the accelerations to get velocity

Start from definitions $a = \dfrac{dv}{dt}$.

Horizontal: $\dfrac{dv_x}{dt}=0 \Rightarrow v_x = u_x = u\cos\theta$ (constant — Why? no horizontal force)

Vertical: $\dfrac{dv_y}{dt}=-g \Rightarrow v_y = u\sin\theta - g t$ (Why minus? up is positive, gravity points down)

Step 2 — Integrate velocity to get position

$x = \int v_x\,dt = (u\cos\theta)\,t$ $y = \int v_y\,dt = (u\sin\theta)\,t - \tfrac12 g t^2$

Step 3 — The trajectory (path) equation

Eliminate $t$. From the x-equation, $t = \dfrac{x}{u\cos\theta}$. Substitute into $y$: $y = u\sin\theta\cdot\frac{x}{u\cos\theta} - \frac12 g\left(\frac{x}{u\cos\theta}\right)^2$ $\boxed{y = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\theta}}$ Why a parabola? It has the form $y = ax - bx^2$ (quadratic in $x$) ⇒ a downward parabola.

Step 4 — Time of flight $T$

The projectile lands when $y=0$ again: $0 = t(u\sin\theta - \tfrac12 g t) \Rightarrow t=0 \text{ or } t = \frac{2u\sin\theta}{g}$ $\boxed{T = \frac{2u\sin\theta}{g}}$ Why this step? $t=0$ is the launch; the other root is the landing. Only vertical motion decides $T$.

Step 5 — Maximum height $H$

At the top, $v_y=0$ ⇒ $u\sin\theta - g t = 0 \Rightarrow t_{\text{top}} = \dfrac{u\sin\theta}{g} = T/2$. Plug into $y$:

= \frac{u^2\sin^2\theta}{g} - \frac{u^2\sin^2\theta}{2g}$$ $$\boxed{H = \frac{u^2\sin^2\theta}{2g}}$$ ### Step 6 — Range $R$ Horizontal distance during the whole flight (constant $v_x$ × total time): $$R = u\cos\theta\cdot T = u\cos\theta\cdot\frac{2u\sin\theta}{g} = \frac{2u^2\sin\theta\cos\theta}{g}$$ Use $2\sin\theta\cos\theta=\sin2\theta$: $$\boxed{R = \frac{u^2\sin2\theta}{g}}$$ **Max range:** $\sin2\theta=1 \Rightarrow 2\theta=90^\circ \Rightarrow \theta=45^\circ$, giving $R_{\max}=u^2/g$. > [!intuition] Complementary angles > $\sin2\theta = \sin(180^\circ-2\theta)$, so $\theta$ and $(90^\circ-\theta)$ give the **same range**. > e.g. $30^\circ$ and $60^\circ$ land in the same spot (but $60^\circ$ flies higher & longer in time). --- ## Worked Examples > [!example] Example 1 — Standard launch > A ball is thrown at $u=20\,\text{m/s}$, $\theta=30^\circ$, $g=10\,\text{m/s}^2$. Find $T$, $H$, $R$. > > - $u_x = 20\cos30^\circ = 17.3\,\text{m/s}$, $u_y = 20\sin30^\circ = 10\,\text{m/s}$. *Why? split into independent pieces.* > - $T = \dfrac{2(10)}{10} = 2\,\text{s}$. *Why? vertical motion sets flight time.* > - $H = \dfrac{(10)^2}{2(10)} = 5\,\text{m}$. *Why? uses only $u_y$.* > - $R = u_x\cdot T = 17.3\times 2 = 34.6\,\text{m}$. *Why? constant horizontal speed × time.* > [!example] Example 2 — Horizontal projectile (the famous "drop vs shoot") > A stone is thrown **horizontally** at $u=15\,\text{m/s}$ from a $20\,\text{m}$ cliff. When/where does it land? > > - Here $\theta=0$: $u_x=15$, $u_y=0$. *Why? launched flat.* > - Vertical: $20 = \tfrac12(10)t^2 \Rightarrow t^2=4 \Rightarrow t=2\,\text{s}$. *Why? falls under gravity exactly like a dropped stone.* > - A stone simply **dropped** also takes $\sqrt{2\cdot20/10}=2\,\text{s}$ → **same time!** (independence confirmed) > - Horizontal range: $x = 15\times 2 = 30\,\text{m}$. > [!example] Example 3 — Forecast-then-Verify > **Forecast:** Two balls, same speed, angles $25^\circ$ and $65^\circ$. Same range? Same time? > **Verify:** $25^\circ+65^\circ=90^\circ$ → complementary → **same range** ✓. > But $T\propto\sin\theta$, and $\sin65^\circ>\sin25^\circ$ → the $65^\circ$ ball stays up **longer**. Forecast partly right! --- ## Common Mistakes (Steel-manned) > [!mistake] "Horizontal velocity decreases as it goes up." > **Why it feels right:** the ball *slows down* near the top, so it seems all motion is dying. > **The fix:** only the **vertical** component shrinks (gravity acts down). $v_x=u\cos\theta$ stays > rock-constant. What you "see slow" is the total speed because $v_y\to0$ at the top. > [!mistake] "At the highest point velocity is zero." > **Why it feels right:** for a ball thrown straight up, $v=0$ at top — true there. > **The fix:** in a projectile only $v_y=0$ at the top; $v_x\neq0$. The ball is still cruising sideways. > [!mistake] "Heavier projectile falls faster / shorter range." > **Why it feels right:** intuition says heavy = drops quicker. > **The fix:** $a_y=-g$ has **no mass** in it. (Ignoring air drag) mass cancels — Galileo's point. > [!mistake] Using $R=u^2\sin2\theta/g$ when launch & landing heights differ. > **Why it feels right:** it's the formula you memorised. > **The fix:** that formula assumes landing at launch height. For a cliff, go back to $x=u_x t$ with $t$ from the **full** vertical equation $y=u_y t-\tfrac12 gt^2$. --- ## Active Recall > [!recall]- Quick self-test (hide answers!) > - What is $a_x$ for a projectile, and WHY? → $0$, because no horizontal force. > - Which component sets time of flight? → vertical ($u_y$). > - Why is the path a parabola? → $y$ is quadratic in $x$. > - Two angles giving equal range? → complementary, $\theta$ & $90^\circ-\theta$. > - Angle for max range? → $45^\circ$. > [!recall]- Feynman: explain to a 12-year-old > Imagine rolling a marble off a table while a friend lets a marble **fall straight down** at the > exact same moment. They smack the floor **together** — clink! — every time. Gravity pulls them > down at the same rate no matter how fast they're zooming sideways. So a thrown ball is really > doing two things at once: gliding forward at a steady pace, and falling like any dropped thing. > Mix those two and you get the curved arc you see — a parabola. > [!mnemonic] Remember the split & formulas > **"Cos Carries, Sin Soars."** > $\cos\theta$ → horizontal (carries it far, gives Range). $\sin\theta$ → vertical (soars up, gives Height & Time). > And **"45 is the max-fly"** for longest range. --- ## Flashcards #flashcards/physics What is the horizontal acceleration of an ideal projectile and why? ::: $a_x=0$ because gravity has no horizontal component; no horizontal force. What are the two parametric position equations? ::: $x=u\cos\theta\,t$ and $y=u\sin\theta\,t-\tfrac12 gt^2$. Why do the horizontal and vertical motions evolve independently? ::: Because Newton's law holds per component and gravity only affects $a_y$; the equations share only time $t$. Trajectory equation of a projectile? ::: $y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}$ (a parabola). Time of flight formula (level ground)? ::: $T=\dfrac{2u\sin\theta}{g}$. Maximum height formula? ::: $H=\dfrac{u^2\sin^2\theta}{2g}$. Range formula and the angle for maximum range? ::: $R=\dfrac{u^2\sin2\theta}{g}$; maximum at $\theta=45^\circ$, $R_{\max}=u^2/g$. Which angles give the same range? ::: Complementary angles $\theta$ and $90^\circ-\theta$. A bullet fired horizontally vs one dropped from same height — which lands first? ::: Both land simultaneously (identical vertical motion). At the top of the trajectory, what is the velocity? ::: Vertical part $v_y=0$, but horizontal $v_x=u\cos\theta$ is unchanged. --- ## Connections - [[Vectors — Resolving into Components]] (why we split into $\cos\theta,\sin\theta$) - [[1-D Kinematics — Equations of Motion]] (each axis is just 1-D motion) - [[Free Fall and g]] (vertical part is free fall) - [[Relative Velocity]] (projectile in a moving frame, e.g. monkey-hunter problem) - [[Newton's Second Law]] (component form gives the independence) - [[Calculus — Integration]] (deriving $v$ and $x$ from $a$) ## 🖼️ Concept Map ```mermaid flowchart TD N["Newton's law component form"] -->|gives| AX["ax = 0"] N -->|gives| AY["ay = -g"] G["Gravity 0,-mg"] -->|only force| N AX -->|integrate| VX["vx = u cosθ constant"] AY -->|integrate| VY["vy = u sinθ - g t"] AX -->|decoupled from| AY VX -->|integrate| X["x = u cosθ t"] VY -->|integrate| Y["y = u sinθ t - ½gt²"] X -->|shared clock t| Y X -->|eliminate t| TRAJ["Trajectory parabola"] Y -->|eliminate t| TRAJ Y -->|set y=0| T["Time of flight T = 2u sinθ / g"] Y -->|set vy=0| H["Max height H"] ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, projectile motion ka core idea bahut simple hai: ek ball jab tum throw karte ho, woh > do alag-alag motions ek saath kar rahi hoti hai. Ek **horizontal** direction mein — yahaan koi > force nahi (gravity sirf neeche khinchti hai), isliye horizontal speed $u\cos\theta$ **constant** > rehti hai. Doosri **vertical** direction mein — yahaan gravity $g$ se neeche pull hota hai, bilkul > free-fall jaise. Dono motions ka sirf ek common cheez share hota hai: **time $t$**. Isi ko hum > "independence" kehte hain. > > Yeh independence kahaan se aaya? Newton ka law $\vec F=m\vec a$ har axis pe alag-alag lagta hai. > Gravity ka horizontal component zero hai, isliye $a_x=0$; aur vertical component $-g$ hai, isliye > $a_y=-g$. Equations ek doosre mein ghuse nahi hue — isliye hum aaram se alag-alag solve kar sakte > hain. Yahi reason hai ki agar tum ek bullet horizontally fire karo aur ek bullet same height se > neeche drop karo, **dono zameen pe ek hi time pe girti hain**. Magic lagता hai, par physics simple hai. > > Formulas ratne ki zaroorat nahi — derive karo. $x=u\cos\theta\,t$ aur $y=u\sin\theta\,t-\tfrac12 gt^2$ > se sab kuch nikal aata hai. $y=0$ rakho toh time of flight $T=2u\sin\theta/g$ mil jaata hai. Top pe > $v_y=0$ rakho toh max height $H=u^2\sin^2\theta/2g$. Aur range $R=u^2\sin2\theta/g$, jiska maximum > $45^\circ$ pe aata hai. Ek mantra yaad rakho: **"Cos carries (range), Sin soars (height & time)."** ![[audio/1.1.19-Projectile-motion-—-horizontal-vertical-independence,-full-derivation.mp3]]