Visual walkthrough — Range, max height, time of flight — all derived

1.1.20 · D2 · Physics › Measurement, Vectors & Kinematics › Range, max height, time of flight — all derived

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Step 1 — Throw draw karo aur jo dikhta hai usse naam do

KYA. Ek ball ground se ek terhi disha mein tezi se nikalti hai. Chaliye picture ko naam dete hain taaki hum uske baare mein baat kar sakein.

• The speed launch ke waqt — ball kitni tezi se haath se nikalti hai — hum ise $u$ kehte hain. Yeh ek number hai, metres per second mein measure hota hai.
• Slant — tum ne kitna upar ki taraf throw kiya, flat ground se upar measure kiya — hum ise $\theta$ (Greek letter "theta", sirf angle ka naam) kehte hain.

KYUN. Kisi bhi maths se pehle, hume words ko marks se jodhna hai. $u$ arrow ki ek length hai; $\theta$ us arrow aur ground ke beech ka opening hai. Aage ka sab kuch sirf inheen dono se banta hai.

PICTURE. Amber arrow dekho: iski length speed $u$ hai, aur amber angle jo iske aur white ground line ke beech hai woh $\theta$ hai. Abhi kuch hila nahi — yeh launch ka pal hai.

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Step 2 — Ek arrow ko do seedhe arrows mein todho

KYA. Terha arrow seedhe samajhna mushkil hai. Hum ise do aisi arrows se replace karte hain jo milkar wahi banaati hain: ek flat (aage) aur ek seedhi upar.

YEH TOOL KYUN — todha kyun? Kyunki gravity hamesha seedha neeche kheenchti hai. Usse aage ki motion se koi matlab nahi. Toh agar hum "aage" aur "upar" alag kar dein, gravity sirf upar wale part ko affect karegi, aur aage wala part trivially simple ho jaata hai (koi cheez use push nahi karti). Yeh split woh ek trick hai jo projectiles ko solve karne laayak banaati hai. Dekho Vectors — resolving into components.

Do naye arrows ki lengths paane ke liye hum ek right triangle use karte hain — ek triangle jisme ek square corner ho. Terhe arrow ko slide karo taaki woh is triangle ki terhi side (hypotenuse) ban jaaye:

• Flat bottom side ki length $u\cos \theta$ hai. Yahan $\cos \theta$ (cosine) matlab hai "slant ka kitna hissa aage point karta hai" — $0$ aur $1$ ke beech ek fraction.
• Vertical side ki length $u\sin \theta$ hai. Yahan $\sin \theta$ (sine) matlab hai "slant ka kitna hissa upar point karta hai".

PICTURE. Amber slant hypotenuse hai; cyan arrow ground ke saath forward speed $u\cos \theta$ hai; white vertical arrow upward speed $u\sin \theta$ hai. Chhota square right angle mark karta hai.

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Step 3 — Up-arrow ko marte aur phir wapas aate dekho (vertical motion)

KYA. Abhi sirf vertical part follow karo. Yeh $u\sin \theta$ se upar jaate shuru hota hai, slow hota hai, top par rukta hai, phir neeche tezi se aata hai.

KYUN. Gravity har second ek fixed amount ki upward speed ghataati hai. Woh fixed amount $g$ hai (Earth ke paas lagbhag $10\text{m/s}$ har second). Dekho Free Fall & Acceleration due to gravity. Kyunki change steady hai, vertical velocity time ke saath ek straight line hai jo neeche jhukti hai.

Kisi bhi time $t$ par vertical velocity (dekho Equations of Motion (1D kinematics)): $v_y=\underset{\text{starting up-speed}}{\underbrace{u\sin \theta }}-\underset{\text{speed gravity has removed by time }t}{\underbrace{gt}}$

• $u\sin \theta$ — woh up-speed jisse hum shuru karte hain.
• $gt$ — $t$ seconds ke baad gravity ne kitni up-speed churayi.
• Minus sign — gravity upward motion ghataati hai.

PICTURE. Graph: vertical velocity ki ek seedhi amber line jo $u\sin \theta$ par positive se shuru hoti hai, zero cross karti hai (peak, cyan dot se mark), phir negative ho jaati hai (gir raha hai). Time white axis ke saath chalta hai.

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Step 4 — Ball kitni der float karti hai yeh nikalo (time of flight)

KYA. Hum $T$ chahte hain, launch se lekar ball ke ground par wapas aane tak ka total time.

YEH METHOD KYUN. Vertical position $0$ se shuru hoti hai aur $0$ par wapas aani chahiye. Toh hum poochhte hain: kin times par height zero hai? Vertical motion se bani height hai: $y=\underset{\text{how far up the throw carries it}}{\underbrace{u\sin \theta t}}-\underset{\text{how far gravity drags it back}}{\underbrace{\frac{1}{2}g{t}^2}}$

$y=0$ set karo (ground par) aur common $t$ factor out karo: $0=t\left(u\sin \theta -\frac{1}{2}gt\right)$

• $t=0$ — launch ka pal khud (haan, woh tab bhi ground par hai). Hum yeh root phenk dete hain.
• $u\sin \theta -\frac{1}{2}gt=0$ — solve karne par landing time milta hai.

$\boxed{T=\frac{2u\sin \theta }{g}}$

• $u\sin \theta$ — zyada upar throw ⇒ zyada lamba flight.
• Denominator mein $g$ — zyada strong gravity ⇒ chhota flight.
• Woh $2$ — upar jaana plus neeche aana, exactly Step 3 wali mirror symmetry.

PICTURE. Height-versus-time curve: ek smooth amber arch jo $0$ se shuru hoti hai, peak karti hai, $0$ par wapas aati hai. Do ground-touch moments $t=0$ aur $t=T$ mark hain; cyan bracket poore span $T$ ko label karta hai.

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Step 5 — Forward-arrow ko time $T$ tak ride karne do (range)

KYA. Ab Step 2 ka forward part wapas laao, $u_x=u\cos \theta$. Koi cheez ise sideways push nahi karti, isliye yeh poore time ball airborne hai is constant speed se chalta hai.

KYUN. Constant speed par distance simply $\text{speed}\times \text{time}$ hai. Available time exactly woh float-time $T$ hai jo humne abhi nikala — yeh $T$ vertical world (jisne clock set ki) aur horizontal world (jisne clock use ki) ke beech ek maatra bridge hai.

$R=u_x\cdot T=\underset{\text{steady forward speed}}{\underbrace{u\cos \theta }}\times \underset{\text{time in air, from Step 4}}{\underbrace{\frac{2u\sin \theta }{g}}}=\frac{2u^2\sin \theta \cos \theta }{g}$

• $u\cos \theta$ — poori flight mein forward speed bani rahi.
• $\frac{2u\sin \theta }{g}$ — woh carrying kitni der chalti hai.
• Multiply karne par $u^2$ milta hai (speed do baar aata hai) aur ek $\sin \theta \cos \theta$ pair.

PICTURE. Ground ke upar poora parabola drawn, neeche ek cyan horizontal bar landing distance $R$ measure karta hai. Chhote clocks/marks forward arrow ko equal ticks mein equal distances step karte dikhate hain (constant speed), jabki height upar neeche hoti hai.

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Step 6 — Double-angle wali safai

KYA. Clumsy $2\sin \theta \cos \theta$ ka ek famous chhota naam hai. Trigonometric identities — double angle se: $2\sin \theta \cos \theta =\sin 2\theta$

ISKO KYUN USE KAREIN. Do reasons. (1) Yeh saaf hai. (2) Yeh best angle turant dikhata hai: ek single sine ko maximise karna zyada aasaan hai. Substitute karne par:

$\boxed{R=\frac{u^2\sin 2\theta }{g}}$

• $u^2$ — do guna tez phenko, chaar guna door jaao.
• $\sin 2\theta$ — poori angle-dependence yahan hai, aur $\sin$ kabhi $1$ se zyada nahi hota.
• $g$ — zyada gravity landing point ko paas kheenchti hai.

PICTURE. $R$ ka $\theta$ ke against plot $0^{\circ }$ se $90^{\circ }$ tak: ek amber hump jo bilkul beech mein $45^{\circ }$ par peak karta hai. Cyan guide lines peak se axis tak girte hain; mirror pair (jaise $30^{\circ }$ aur $60^{\circ }$) equal heights par mark hain.

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Step 7 — Edge cases (reader ko kabhi map se girne mat do)

KYA AUR KYUN. Ek formula jis par tum bharosa karo use apni extremes survive karni chahiye. Hum corners test karte hain.

Case $\theta =0^{\circ }$ (flat ground ke saath throw kiya). Tab $\sin 0^{\circ }=0$, isliye $R=0$. Sahi hai: koi upward part nahi hone se, vertical motion $T=\frac{2u\sin 0}{g}=0$ deta hai — ball already ground par hai aur kabhi nahi uthi. Air mein zero time ⇒ zero distance.

Case $\theta =90^{\circ }$ (seedha upar throw kiya). Tab $\sin 2\theta =\sin 180^{\circ }=0$, isliye $R=0$. Sahi hai: woh upar jaati hai aur seedha tumhare sir par wapas aati hai. Sabse lamba flight time, lekin zero forward travel kyunki $u\cos 90^{\circ }=0$.

Extremes ke beech. $R$ $0$ se badhta hai, $45^{\circ }$ par peak karta hai, aur wapas $0$ par aata hai — Step 6 ka poora hump. Koi angle unexplained nahi rehta.

Warning — same-height sirf. Yahan har box ne assume kiya ki launch aur landing same level par hain. Cliff se phenko toh $y$ wapas $0$ par nahi aata; tumhe drop include karke Step 4 ka quadratic phir se solve karna hoga.

PICTURE. Teen chhoti trajectories side by side: flat $\theta =0$ streak jo kabhi nahi uthi, vertical $\theta =90^{\circ }$ spike jo launch point par wapas aati hai, aur graceful $45^{\circ }$ arch jo sabse door jaati hai — sab ek baseline share karte hain.

Recall Quick self-test

Range $0^{\circ }$ aur $90^{\circ }$ dono par zero kyun hai? ::: $0^{\circ }$ par koi air time nahi ($T=0$); $90^{\circ }$ par koi forward speed nahi ($u\cos 90^{\circ }=0$). Range ke dono drivers ends par zero ho jaate hain. Kaun si do cheezein multiply hokar range deti hain? ::: Constant forward speed $u\cos \theta$ times air time $T=\frac{2u\sin \theta }{g}$.

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Ek-picture summary

PICTURE. Sab kuch ek blueprint par: launch arrow $u\cos \theta$ (cyan, flat) aur $u\sin \theta$ (white, up) mein split; vertical part clock $T=\frac{2u\sin \theta }{g}$ drive karta hai; forward part us clock par ride karke amber parabola trace karta hai; landing distance $R=\frac{u^2\sin 2\theta }{g}$ neeche bracket mein; peak height $H$ upar label par. Ek nazar, ek derivation.

Recall Feynman retelling — poori walk simple words mein (click)

Tum ek ball terhi disha mein phente ho. Pehli trick: throw ko do mein kaato — ek part aage jaata hai, ek part seedha upar jaata hai. Gravity aalsi hai: woh sirf up-part se ladhti hai. Toh up-part apni ek chhoti si story hai — woh chadta hai, slow hota hai, rukta hai, aur girta hai, ek certain amount of time $T$ leta hua. Kyunki slow-down bilkul steady hai, upar jaane ka time neeche aane ke time ke barabar hai, aur yahi se "times two" aata hai: $T=\frac{2u\sin \theta }{g}$.

Wazahat forward part sirf cruise karta rehta hai — kuch bhi ise slow nahi karta — ek fixed speed $u\cos \theta$ par. Toh jitni door woh land karta hai woh simply cruise speed × float time hai. Unhe multiply karo aur tumhe $\frac{2u^2\sin \theta \cos \theta }{g}$ milta hai, jise trig $\frac{u^2\sin 2\theta }{g}$ rename karne deta hai.

Kyunki $\sin$ andar $90^{\circ }$ par sabse bada hota hai, aur woh $90^{\circ }$ do guna launch angle hai, best throw $45^{\circ }$ hai — kaafi der tak upar rehne aur aage tezi se drive karne ke beech ek fair balance. Flat throw karo aur tum fast ho lekin jaldi land karo; zyada steep throw karo aur tum long float karo lekin barely aage badhte ho. Aur do extremes par — dead flat ya seedha upar — tum kahin nahi jaate.

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Connections

• Range, max height, time of flight — all derived — parent, jahan boxed results rehte hain.
• Vectors — resolving into components — Step 2 ki arrow-splitting.
• Projectile Motion — components & independence — kyun forward aur up motions aapas mein baat nahi karte.
• Equations of Motion (1D kinematics) — $v_y$ aur $y$ formulas jo Steps 3–4 mein use hue.
• Free Fall & Acceleration due to gravity — $g$ kahan se aata hai.
• Trigonometric identities — double angle — Step 6 identity.
• Energy method — max height via conservation — $H$ tak pahunchne ka ek alternate route.