Projectile motion — horizontal - vertical independence, full derivation: Visual walkthrough

1.1.19 · D2 · Physics › Measurement, Vectors & Kinematics › Projectile motion — horizontal - vertical independence, full

Definition Setup aur uske assumptions (pehle padho)

Koi bhi symbol aane se pehle, scene set karo:

Launch point = origin. Hum sab kuch wahan se measure karte hain jahan ball haath se chhootti hai.
$x$ — horizontal displacement: ball launch point se kitna sideways gayi hai, metres mein. Launch par $x = 0$ .
$y$ — vertical displacement: ball launch point se kitna upar hai (ya agar negative ho toh neeche), metres mein. Launch par $y = 0$ .
$t$ — launch ke baad beeta hua time, seconds mein, jahan $t = 0$ woh instant hai jab ball haath chhoddti hai.
Assumption 1 — no air resistance. Flight ke dauran sirf gravity hi kaam karti hai.
Assumption 2 — constant $g$ . Gravity ki strength puri flight mein nahi badlti (chhoti heights ke liye sach hai jinka hum deal karte hain).
Assumption 3 — level ground. Launch aur landing ek hi height par hoti hai ( $y = 0$ ). Step 9 dikhata hai jab yeh fail ho (ek cliff par) toh kya badalna padta hai.

Step 1 — Ek arrow, do shadows

HUM KYA karte hain: Hum page ke corner se ("origin" — woh point jahan horizontal ruler aur vertical ruler dono zero padhte hain) ek ball fenkte hain. Yeh speed $u$ se nikalti hai — metres per second mein kitni fast — flat ground se upar angle $θ$ par.

HUM YEH KYU karte hain: Ek tilke hue arrow ke baare mein seedha sochna mushkil hai. Lekin ek tilka hua arrow secretly do simpler arrows rakhta hai: iska kitna hissa sideways point karta hai aur kitna upar. Agar hum ise split kar sakein, toh hum har piece ko plain 1-D motion se handle kar sakte hain (dekho Vectors — Resolving into Components).

PICTURE: Figure dekho. Red arrow launch velocity hai. Uski tip se horizontal ruler tak ek seedhi line drop karo — us shadow ki length horizontal piece hai. Vertical ruler ki taraf slide karo — woh shadow vertical piece hai. Arrow, uske do shadows, aur corner milke ek right triangle banate hain.

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

Horizontal ke liye cosine aur vertical ke liye sine kyun? Kyunki launch triangle mein ground side angle ke next (adjacent) baithti hai, aur height side uske saamne (opposite) baithti hai. Yahi poori wajah hai — koi formula yaad nahi karna, bas triangle padho.

Step 2 — Do alag kahaniyaan, ek shared clock

HUM KYA karte hain: Hum poochhte hain: jab ball ud rahi hoti hai, toh use kya push karta hai? Sirf gravity, seedha neeche $g$ strength se (Earth ke paas lagbhag $10 m/s^{2}$ — dekho Free Fall and g). Sideways push karne wala kuch nahi hai.

HUM YEH KYU karte hain: Newton's Second Law ko har direction par apply karne par ( $F_{x} = m a_{x}$ , $F_{y} = m a_{y}$ ), ek force jo sirf neeche point karti hai woh sirf up-down motion ko badal sakti hai. Sideways motion par koi force nahi hai, toh kuch nahi badalta.

PICTURE: Red gravity arrow purely neeche point karta hai. Horizontal ruler par uska shadow zero length ka hai — yeh visual statement hai "gravity ka koi sideways part nahi hai". Do side panels do independent kahaniyaan dikhate hain jo ek hi stopwatch $t$ (launch ke baad beeta hua time) share karti hain.

Do kahaniyaon ko jodhne wali sirf ek cheez hai $t$ , ek clock jo dono ke liye tick karti hai.

Step 3 — Acceleration se velocity tak (ek integration)

HUM KYA karte hain: Acceleration woh rate hai jis par velocity change hoti hai. "Velocity kitni fast change hoti hai" se wapas "velocity khud" par jaane ke liye hum elapsed time $t$ par un sab chhoti changes ko jodthe hain — yeh jodhna integration hai (dekho Calculus — Integration).

Integration kyun aur kuch kyun nahi? Acceleration "change per second" batata hai. Velocity us change ka running total hai. Kisi quantity ko uski rate se recover karna bilkul wohi hai jo integration karta hai. Koi aur cheez rate se quantity recover nahi kar sakti.

PICTURE: Velocity versus time ke do graphs. Horizontal velocity ek flat red line hai — yeh kabhi nahi hilti. Vertical velocity ek red line hai jo neeche slope karti hai — yeh $u sin θ$ se shuru hoti hai aur har second $g$ se girती hai, peak par zero cross karti hai, phir negative ho jaati hai (girna).

Step 4 — Velocity se position tak (dobara integrate karo)

HUM KYA karte hain: Velocity woh rate hai jis par position change hoti hai. Ise dobara elapsed time $t$ par jodho aur hum jaante hain ball actually kahan hai — launch point se uska horizontal displacement $x$ aur vertical displacement $y$ (dono upar setup box mein define kiye gaye hain).

KYU: Step 3 jaisi hi logic, ek level upar. Position velocity ka running total hai, toh hum ek baar phir integrate karte hain. (Dekho 1-D Kinematics — Equations of Motion — yeh exactly wohi equations hain, har axis ke liye ek.)

Do integrations, explicitly:

Horizontal velocity $v_{x} = u cos θ$ constant hai, toh time $t$ par uska running total bas constant $\times$ time hai: $x = (u cos θ) t$ .
Vertical velocity $v_{y} = u sin θ - g t$ ke do pieces hain. Constant part $u sin θ$ ko time par integrate karne se $(u sin θ) t$ milta hai. Badhte hue part $- g t$ ko integrate karne se $- \frac{1}{2} g t^{2}$ milta hai — yahi se $\frac{1}{2}$ aata hai: kisi quantity ka total jo $0$ se $g t$ tak linearly badhti hai woh average value $\frac{1}{2} g t$ times time $t$ hai, yani $\frac{1}{2} g t^{2}$ (jo picture mein us triangle ka area hai).

PICTURE: Velocity-time graphs par, har line ke neeche shaded red area wohi distance hai jo travel ki gayi. Flat line ka area ek rectangle hai → steadily badhta hai. Sloped line ka area ek triangle hai → classic $- \frac{1}{2} g t^{2}$ falling term.

Neeche sab kuch in do lines se nikala jaata hai.

Step 5 — Clock mitao: parabola appear hoti hai

HUM KYA karte hain: Hum flight ki shape chahte hain — vertical displacement $y$ sideways distance $x$ ke function ke roop mein, time chhupa ke. Toh hum $x$ -equation ko $t$ ke liye solve karte hain aur $y$ -equation mein substitute karte hain.

KYU: Arc ki photograph mein koi clock nahi hoti. Us photo se match karne ke liye hum $t$ ko eliminate karke pure $y$ -versus- $x$ relationship banana chahte hain.

PICTURE: Red curve flight hai. Notice karo iska algebraic form $y = (kuch) x - (kuch) x^{2}$ hai — ek term linear in $x$ ise upar kheenchti hai, aur ek term $x^{2}$ mein ise neeche. Yeh ek downward parabola ki signature hai.

Step 6 — Yeh kab land karta hai? Time of flight $T$

HUM KYA karte hain: "Level ground par landing" matlab height wapas zero ho jaati hai, $y = 0$ . Hum $y = 0$ ko time ke liye solve karte hain.

KYU: Sirf vertical story decide karti hai kab ball ground touch karti hai — sideways motion ise kabhi ground par wapas nahi laati. Toh hum $y$ -equation ko zero set karte hain. (Yeh Assumption 3 use karta hai — level ground.)

PICTURE: Parabola ground line ko do red dots par cross karti hai: ek start par ( $t = 0$ ) aur ek landing par ( $t = T$ ). In dono ke beech peak bilkul halfway baithti hai.

Step 7 — Kitna upar? Maximum height $H$

HUM KYA karte hain: Bilkul upar ball ka uthna ruk jaata hai: uski upward speed momentarily zero ho jaati hai. $v_{y} = 0$ set karo, woh instant dhundho, $y$ mein plug karo.

KYU: Peak woh ek jagah hai jahan $v_{y}$ zero cross karta hai (upar se neeche). Woh ek condition upar ko pinpoint karti hai.

PICTURE: Apex par vertical red arrow kuch nahi raha, jabki horizontal red arrow abhi bhi full-length hai — ball abhi bhi sideways cruise kar rahi hai bilkul upar par bhi.

Step 8 — Kitni door? Range $R$

HUM KYA karte hain: Ball poori flight mein constant speed se sideways drift karti hai, toh range = sideways speed $\times$ total time.

KYU: Horizontal speed kabhi nahi badlti (Step 2), toh distance simply speed times time hai — physics mein sabse aasaan multiplication.

PICTURE: Arc ke neeche $R$ length ki ek red horizontal bar. Uske upar, ek chhota dial dikhata hai $R$ bilkul $θ = 4 5^{\circ}$ par sabse bade value tak phool rahi hai, phir dobara shrink ho rahi hai.

Step 9 — Edge cases (inhe kabhi skip mat karo)

HUM KYA karte hain: Hum derivation ko uske extreme angles par test karte hain, aur Assumption 3 ko lift karte hain dekhne ke liye kya badalta hai.

KYU: Ek rule jise tumne sirf "nice" case mein test kiya woh ek aise rule nahi jis par tum trust karo. Corners check karo.

PICTURE: Teen chhoti red flights side by side: flat fire ki gayi, seedha upar fire ki gayi, aur balanced $4 5^{\circ}$ throw.

Ek-picture summary

Har step, ek diagram: red arc, uske do shadow-motions, landing dots, peak, aur range bar — poori derivation compress ki gayi.

Recall Feynman retelling — plain words mein poora walkthrough

Ek ball phenko. Woh ek tilka hua push actually do pushes in disguise hai: ek sideways aur ek upward — arrow ke do shadows. Ab chhhod do. Gravity sirf neeche kheenchti hai, toh yeh sideways shadow ko kabhi nahi chhooti — ball steadily aage glideti rehti hai hamesha ke liye. Jabki upward shadow gravity se kha raha hai: yeh slow hoti hai, top par ruk jaati hai, phir tezi se wapas girती hai. "Steady forward" ko "rise-then-fall" mein jodo aur tum ek smooth arch draw karte ho — ek parabola. Yeh kab land karti hai? Jab up-down story ground par wapas aati hai — woh time of flight hai, aur yeh sirf upward shadow se set hoti hai. Kitna upar? Jahan upward speed zero ho jaati hai. Kitni door? Steady sideways speed times total time — sabse zyada jab throw ko forty-five degrees par evenly split karo. Aur agar tum ise flat fire karo ek cliff se, toh yeh ek ball ke saath ek hi step mein girती hai jise tum simply drop karte ho — clink, saath mein — kyunki unki up-down kahaniyaan bilkul ek hi kahani hain.

Recall Khud rebuild karo

Kaunsa integration acceleration se velocity deta hai? ::: Pehla wala; velocity acceleration ka running total hai. Time of flight dhundhne ki condition kya hai? ::: $y = 0$ set karo (launch height par wapas). Peak dhundhne ki condition kya hai? ::: $v_{y} = 0$ set karo. Path parabola kyun hai? ::: $t$ erase karne ke baad, $y$ $x$ mein quadratic ho jaata hai. Kaunsa angle range maximize karta hai aur kyun? ::: $4 5^{\circ}$ , kyunki wahan $sin 2 θ = 1$ hota hai.