Projectile motion — horizontal - vertical independence, full derivation

1.1.19 · Physics › Measurement, Vectors & Kinematics

Independence kyun hold karta hai? (first principles se)

ISKA MATLAB: $a_{x} = 0$ ⇒ horizontal velocity kabhi nahi badlegi. $a_{y} = - g$ ⇒ vertical velocity bilkul waise hi badlti hai jaise ek drop kiya hua object. Na hi equation mein doosra variable aata hai — ye decoupled hain.

HUME KAISE FAIDA HOGA: har axis ko alag se 1-D kinematics se solve karo, phir shared $t$ se combine karo.

Scratch se puri derivation

Origin se speed $u$ ke saath angle $θ$ ऊपर horizontal se launch karo. Initial velocity ko split karo: $u_{x} = u cos θ, u_{y} = u sin θ$

Step 1 — Accelerations ko integrate karke velocity nikalo

Definitions $a = \frac{d v}{d t}$ se shuru karo.

Horizontal: $\frac{d v _{x}}{d t} = 0 \Rightarrow v_{x} = u_{x} = u cos θ$ (constant — Kyun? koi horizontal force nahi)

Vertical: $\frac{d v _{y}}{d t} = - g \Rightarrow v_{y} = u sin θ - g t$ (minus kyun? upar positive hai, gravity neeche point karti hai)

Step 2 — Velocity ko integrate karke position nikalo

$x = \int v_{x} d t = (u cos θ) t$ $y = \int v_{y} d t = (u sin θ) t - \frac{1}{2} g t^{2}$

Step 3 — Trajectory (path) equation

$t$ eliminate karo. X-equation se, $t = \frac{x}{u cos θ}$ . $y$ mein substitute karo: $y = u sin θ \cdot \frac{x}{u c o s θ} - \frac{1}{2} g (\frac{x}{u c o s θ})^{2}$ $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}$ Parabola kyun hai? Iska form hai $y = a x - b x^{2}$ ( $x$ mein quadratic) ⇒ ek neeche wali parabola.

Figure — Projectile motion — horizontal - vertical independence, full derivation

Step 4 — Time of flight $T$

Projectile tab land karta hai jab $y = 0$ phir se ho: $0 = t (u sin θ - \frac{1}{2} g t) \Rightarrow t = 0 or t = \frac{2 u s i n θ}{g}$ $T = \frac{2 u sin θ}{g}$ Ye step kyun? $t = 0$ launch hai; doosri root landing hai. Sirf vertical motion $T$ decide karta hai.

Step 5 — Maximum height $H$

Top par, $v_{y} = 0$ ⇒ $u sin θ - g t = 0 \Rightarrow t_{top} = \frac{u sin θ}{g} = T /2$ . $y$ mein plug karo:

= \frac{u^2\sin^2\theta}{g} - \frac{u^2\sin^2\theta}{2g}$$ $$\boxed{H = \frac{u^2\sin^2\theta}{2g}}$$ ### Step 6 — Range $R$ Pure flight ke dauran horizontal distance (constant $v_x$ × total time): $$R = u\cos\theta\cdot T = u\cos\theta\cdot\frac{2u\sin\theta}{g} = \frac{2u^2\sin\theta\cos\theta}{g}$$ $2\sin\theta\cos\theta=\sin2\theta$ use karo: $$\boxed{R = \frac{u^2\sin2\theta}{g}}$$ **Max range:** $\sin2\theta=1 \Rightarrow 2\theta=90^\circ \Rightarrow \theta=45^\circ$, jisse $R_{\max}=u^2/g$ milta hai. > [!intuition] Complementary angles > $\sin2\theta = \sin(180^\circ-2\theta)$, isliye $\theta$ aur $(90^\circ-\theta)$ **same range** dete hain. > Jaise $30^\circ$ aur $60^\circ$ same jagah land karte hain (lekin $60^\circ$ wala zyada oopar jaata hai aur time mein zyada rehta hai). --- ## Worked Examples > [!example] Example 1 — Standard launch > Ek ball $u=20\,\text{m/s}$, $\theta=30^\circ$, $g=10\,\text{m/s}^2$ se throw ki jaati hai. $T$, $H$, $R$ nikalo. > > - $u_x = 20\cos30^\circ = 17.3\,\text{m/s}$, $u_y = 20\sin30^\circ = 10\,\text{m/s}$. *Kyun? independent pieces mein split karo.* > - $T = \dfrac{2(10)}{10} = 2\,\text{s}$. *Kyun? vertical motion flight time set karta hai.* > - $H = \dfrac{(10)^2}{2(10)} = 5\,\text{m}$. *Kyun? sirf $u_y$ use karta hai.* > - $R = u_x\cdot T = 17.3\times 2 = 34.6\,\text{m}$. *Kyun? constant horizontal speed × time.* > [!example] Example 2 — Horizontal projectile (famous "drop vs shoot") > Ek stone **horizontally** $u=15\,\text{m/s}$ se $20\,\text{m}$ ki cliff se throw kiya jaata hai. Kab/kahan land karega? > > - Yahan $\theta=0$: $u_x=15$, $u_y=0$. *Kyun? flat launch kiya.* > - Vertical: $20 = \tfrac12(10)t^2 \Rightarrow t^2=4 \Rightarrow t=2\,\text{s}$. *Kyun? bilkul dropped stone ki tarah gravity ke under girta hai.* > - Simply **drop** kiya gaya stone bhi $\sqrt{2\cdot20/10}=2\,\text{s}$ leta hai → **same time!** (independence confirmed) > - Horizontal range: $x = 15\times 2 = 30\,\text{m}$. > [!example] Example 3 — Forecast-then-Verify > **Forecast:** Do balls, same speed, angles $25^\circ$ aur $65^\circ$. Same range? Same time? > **Verify:** $25^\circ+65^\circ=90^\circ$ → complementary → **same range** ✓. > Lekin $T\propto\sin\theta$, aur $\sin65^\circ>\sin25^\circ$ → $65^\circ$ wali ball **zyada der** tak hawa mein rehti hai. Forecast aadha sahi tha! --- ## Common Mistakes (Steel-manned) > [!mistake] "Horizontal velocity upar jaate waqt kam ho jaati hai." > **Ye sahi kyun lagta hai:** ball top ke paas *slow down* hoti lagti hai, toh lagta hai saari motion khatam ho rahi hai. > **Fix:** sirf **vertical** component ghatta hai (gravity neeche act karta hai). $v_x=u\cos\theta$ bilkul > constant rehti hai. Jo tum "slow" dekhte ho wo total speed hai kyunki $v_y\to0$ top par. > [!mistake] "Sabse oopar point par velocity zero hoti hai." > **Ye sahi kyun lagta hai:** seedha upar throw kiye ball ke liye, top par $v=0$ hota hai — wahan ye sach hai. > **Fix:** projectile mein sirf $v_y=0$ hota hai top par; $v_x\neq0$. Ball sideways chalti rehti hai. > [!mistake] "Bhaari projectile zyada tezi se girta hai / kam range milti hai." > **Ye sahi kyun lagta hai:** intuition kehta hai heavy = jaldi girta hai. > **Fix:** $a_y=-g$ mein **koi mass nahi** hai. (Air drag ignore karo) mass cancel ho jaata hai — Galileo ka point. > [!mistake] $R=u^2\sin2\theta/g$ use karna jab launch aur landing heights alag hon. > **Ye sahi kyun lagta hai:** ye formula yaad kiya hua hai. > **Fix:** wo formula assume karta hai ki landing launch height par hi ho. Cliff ke liye, wapas $x=u_x t$ par jao aur $t$ **poori** vertical equation $y=u_y t-\tfrac12 gt^2$ se nikalo. --- ## Active Recall > [!recall]- Quick self-test (answers chhupa lo!) > - Projectile ke liye $a_x$ kya hai, aur KYUN? → $0$, kyunki koi horizontal force nahi. > - Kaunsa component time of flight set karta hai? → vertical ($u_y$). > - Path parabola kyun hai? → $y$, $x$ mein quadratic hai. > - Do angles jo equal range dete hain? → complementary, $\theta$ aur $90^\circ-\theta$. > - Max range ke liye angle? → $45^\circ$. > [!recall]- Feynman: ek 12-saal ke bacche ko explain karo > Socho ek marble table se roll kar raha hai aur ek dost us exact same waqt ek marble **seedha neeche girata** hai. > Wo dono floor par **saath mein** girate hain — clink! — har baar. Gravity unhe neeche > same rate se kheenchti hai chahe wo sideways kitni bhi tezi se ja rahe hon. Isliye ek throw ki gayi ball actually > ek saath do kaam kar rahi hai: ek steady pace se aage glide karna, aur kisi bhi dropped cheez ki tarah girna. > In dono ko milao aur tum wo curved arc dekhte ho — ek parabola. > [!mnemonic] Split aur formulas yaad karo > **"Cos Carries, Sin Soars."** > $\cos\theta$ → horizontal (door le jaata hai, Range deta hai). $\sin\theta$ → vertical (oopar soar karta hai, Height aur Time deta hai). > Aur **"45 is the max-fly"** sabse zyada range ke liye. --- ## Flashcards #flashcards/physics Ek ideal projectile ka horizontal acceleration kya hai aur kyun? ::: $a_x=0$ kyunki gravity ka koi horizontal component nahi hai; koi horizontal force nahi. Do parametric position equations kya hain? ::: $x=u\cos\theta\,t$ aur $y=u\sin\theta\,t-\tfrac12 gt^2$. Horizontal aur vertical motions independently kyun evolve karte hain? ::: Kyunki Newton's law har component ke liye alag hold karta hai aur gravity sirf $a_y$ ko affect karti hai; equations sirf time $t$ share karte hain. Projectile ki trajectory equation kya hai? ::: $y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}$ (ek parabola). Time of flight formula (level ground par)? ::: $T=\dfrac{2u\sin\theta}{g}$. Maximum height formula? ::: $H=\dfrac{u^2\sin^2\theta}{2g}$. Range formula aur maximum range ke liye angle? ::: $R=\dfrac{u^2\sin2\theta}{g}$; maximum $\theta=45^\circ$ par, $R_{\max}=u^2/g$. Kaunse angles same range dete hain? ::: Complementary angles $\theta$ aur $90^\circ-\theta$. Horizontally fire ki gayi bullet vs same height se drop ki gayi bullet — pehle kaun land karti hai? ::: Dono ek saath land karti hain (identical vertical motion). Trajectory ke top par velocity kya hoti hai? ::: Vertical part $v_y=0$, lekin horizontal $v_x=u\cos\theta$ unchanged rehti hai. --- ## Connections - [[Vectors — Resolving into Components]] (isliye hum $\cos\theta,\sin\theta$ mein split karte hain) - [[1-D Kinematics — Equations of Motion]] (har axis sirf 1-D motion hai) - [[Free Fall and g]] (vertical part free fall hai) - [[Relative Velocity]] (moving frame mein projectile, jaise monkey-hunter problem) - [[Newton's Second Law]] (component form independence deta hai) - [[Calculus — Integration]] ($a$ se $v$ aur $x$ derive karna) ## 🖼️ 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"] ```