Free fall — g = 9.8 m - s², sign conventions: Visual walkthrough

1.1.17 · D2 · Physics › Measurement, Vectors & Kinematics › Free fall — g = 9.8 m - s², sign conventions

Step 1 — Pehle number line draw karo, physics se pehle

KYA. Girne ki baat karne se pehle, hum ek vertical ruler draw karte hain. Hum choose karte hain upar = positive. Ground aur sky ab sirf numbers hain: shuruat se 5 metre upar ka spot $+ 5$ hai, shuruat se 3 metre neeche ka spot $- 3$ hai.

KYUN. Free fall mein har " $+$ " aur " $-$ " ek choice hai jo hum ek baar karte hain. Agar hum ruler ko abhi pin down nahi karte, toh baad mein signs jaadu jaisi lagti hain. Ye jaadu nahi hai — ye is ek arrow ka consequence hai.

PICTURE. Figure mein, purple arrow woh direction dikhata hai jise hum positive kehte hain. Jo bhi us ke saath point karta hai woh $+$ count hota hai; jo bhi us ke against point karta hai woh $-$ count hota hai. Notice karo ki gravity ka chhota coral arrow humare positive direction ke against point karta hai. Ye yaad rakho — yahi ek wajah hai ki $g$ minus sign pahnega.

Figure — Free fall — g = 9.8 m - s², sign conventions

Step 2 — "Acceleration" ka matlab: velocity ka slope

KYA. Hum claim karte hain ki free fall mein constant acceleration hoti hai: velocity har second same amount se change hoti hai. Earth ke paas woh amount ka size $g \approx 9.8 m/s^{2}$ hai, direction down hai. Hamare up-positive ruler par ye banata hai $a = - g = - 9.8 m/s^{2} .$

YE NUMBER KYUN, CONSTANT KYUN? Newton's Second Law se: gravity $m g$ force se kheenchti hai, aur $F = ma$ deta hai $ma = m g$ , toh $a = g$ — mass cancel ho jata hai, aur us equation mein kuch bhi time par depend nahi karta, toh $a$ kabhi change nahi hota. Ek value, hamesha (jab tak kuch ground se nahi takrata ya hawa involve nahi hoti — dekho Air Resistance and Terminal Velocity).

PICTURE. Velocity $v$ ko time $t$ ke against plot karo. Kyunki $a$ constant hai, ye graph ek straight line hai. Uski steepness hi acceleration hai. Neeche hum $\frac{d v}{d t}$ ko exactly "steepness of the $v$ – $t$ line" ka matlab dete hain — woh amount jo $v$ har second gain karta hai.

$a = steepness of the v - t line \frac{d v}{d t} = - g$

Step 3 — Slope se wapas velocity tak: pehli equation

KYA. Hum kisi bhi time $t$ par $v$ ka formula chahte hain. Hum jaante hain ki line $v_{0}$ height par shuru hoti hai (initial velocity) aur slope $a = - g$ hai. Ek straight line hoti hai start value + slope × run: $v = line ki starting height v_{0} + slope a time along run t = v_{0} - g t .$

KYUN. Ye sirf graph se straight line padhna hai — ek baar jab hum dekh lete hain ki ye ek line hai toh calculus ki zaroorat nahi. (Parent ka integral $\int d v = \int a d t$ bilkul same baat kehta hai: equal chhoti velocity-gains ko jodna slope × time deta hai.)

PICTURE. Coral line ko neeche follow karo. Har second ye $9.8$ se drop karta hai. Ye positive se shuru hota hai (upar fenka gaya), zero cross karta hai (peak — ball momentarily still hai), phir negative ho jata hai (wapas girna). Ek straight line poori upar-neeche ki story contain karti hai.

Step 4 — Velocity se position tak: line ke neeche ka area

KYA. Position change hoti hai kyunki velocity hume ruler par move karti hai. Displacement (kitna $y$ move kiya) $v$ – $t$ line aur time axis ke beech ke area ke barabar hai.

AREA KYUN? Distance = speed × time. Agar speed constant hoti, toh ye ek rectangle hota. Yahan speed ek straight line ke saath change hoti hai, toh region ek rectangle hai (from $v_{0}$ ) plus ek triangle (changing part se). Shapes ka area jo hum jaante hain — yahi wajah hai ki yahan tool "area" hai, kuch fancy nahi.

PICTURE. Mint rectangle ki height $v_{0}$ aur width $t$ hai: area $v_{0} t$ . Lavender triangle ki width $t$ aur height $g t$ hai (kitni velocity drop hui): area $\frac{1}{2} (t) (g t) = \frac{1}{2} g t^{2}$ . Kyunki triangle neeche girte waqt axis ke neeche baithta hai, ye subtract hota hai.

$y - y_{0} = rectangle v_{0} t - triangle \frac{1}{2} g t^{2}$

Step 5 — Time hatana: speed–distance relation

KYA. Kabhi kabhi hume $t$ ki parwaah nahi hoti, sirf ye jaanna hota hai ki kitni dur ke baad kitni tezi. Hum $t$ ko dono boxed equations se hata dete hain.

KYUN. "20 m girne ke baad impact speed" jaisi problems distance deti hain, time nahi. Ek formula jo $v$ aur displacement $Δ y$ ko directly link karta hai, hume pehle $t$ solve karne se bachata hai.

KAISE (algebra, words mein). Equation 1 se, $t = \frac{v _{0} - v}{g}$ . Use Equation 2 mein daalo aur $t$ sab collapse ho jata hai. Clean result: $v^{2} = v_{0}^{2} - 2 g Δ y, Δ y = y - y_{0} .$

PICTURE. Ise ruler par khichha hua ek energy ledger samjho: $Δ y$ se upar jaana " $v^{2}$ " ka $2 g Δ y$ cost karta hai; girna use refund karta hai. Height aur speed-squared ek-ke-badle-ek trade hote hain.

Step 6 — Peak, aur kyun wahan acceleration zero nahi hoti

KYA. Equation 1 mein $v = 0$ set karo: object momentarily still hai. $0 = v_{0} - g t$ solve karo to pao time to the top: $t_{up} = \frac{v _{0}}{g} .$ Equation 3 mein $v = 0$ ke saath plug karo: max height $H = \frac{v _{0}^{2}}{2 g} .$

YE KYUN MATTER KARTA HAI. Beginners sochte hain "ruka hua = koi acceleration nahi." Lekin $v$ – $t$ line par, peak woh single point hai jahan line zero cross karti hai — wahan line ka slope abhi bhi $- g$ hai. Velocity zero se guzarti hai; acceleration bilkul unchanged chalti rehti hai.

PICTURE. Coral line time-axis ko $t_{up}$ par touch karti hai (velocity zero) lekin apna downward slant jaari rakhti hai. Height curve $y (t)$ usi instant apni rounded summit par pahunchti hai — wahan flat, lekin poore time curve down ho rahi hai.

Step 7 — Ek picture par saare cases

KYA. Wohi teen equations har launch cover karti hain. Sirf $v_{0}$ ka sign aur starting height change hoti hai.

INHE SAATH KYUN DIKHAYEIN. Taaki tum kisi aisi scenario se kabhi na milo jo derivation ne miss ki ho:

Case	$v_{0}$	Line kya karti hai
Rest se drop kiya	$0$	axis par shuru hoti hai, sirf girti hai
Upar fenka	$+$	positive se shuru, peak par zero cross, phir negative
Neeche fenka	$-$	negative se shuru, aur negative hoti jaati hai

PICTURE. Teen velocity lines, same slope $- g$ (sab parallel!), alag starting heights $v_{0}$ . Parallel = har object aur har throw ke liye same acceleration — ek nazar mein free fall ki gehri sachai. Aur degenerate case (dropped, $v_{0} = 0$ ) simply woh line hai jo axis par shuru hoti hai.

Ek-picture summary

KYA. Ek straight $v$ – $t$ line sab kuch generate karti hai: uski starting height $v_{0}$ hai, uska slope $- g$ hai (Equation 1), uska area displacement hai (Equation 2), aur uska zero-crossing peak hai. $t$ hatao aur Equation 3 milti hai. Poora chapter ek line aur us ke neeche ke region mein compress ho gaya.

Recall Feynman retelling — poora walkthrough simple words mein

Pehle maine ek ruler khada karke kaha "upar plus direction hai." Gravity doosri taraf point karti hai, toh ye hamesha minus ke roop mein aati hai. Phir maine notice kiya ki gravity girte hue cheez ko har ek second mein same 9.8 se speed up karti hai — toh agar main plot karoon ki ye kitni tezi se ja rahi hai time ke against, mujhe ek bilkul straight, tilted line milti hai. Ek straight line hai "jahan tum shuru hue plus tumhara slope times the time," aur ye meri pehli equation hai, $v = v_{0} - g t$ . Ye pata lagane ke liye ki cheez kahan hai, maine us line ke neeche trapped area measure kiya — ek rectangle plus ek triangle — aur nikla $y = y_{0} + v_{0} t - \frac{1}{2} g t^{2}$ . Jab mujhe sirf kuch distance ke baad speed ki parwaah thi aur clock ki nahi, maine algebraically time delete kar diya aur mila $v^{2} = v_{0}^{2} - 2 g Δ y$ . Speed ko zero set karne se throw ka top mila — lekin line ka tilt wahan kabhi flat nahi hua, yahi exact wajah hai ki gravity kheenchti rehti hai aur cheez wapas neeche aati hai. Aakhir mein maine ek saath teen throws draw kiye: dropped, up, down — teen parallel lines, same tilt — ek picture of the fact ki sab kuch same girata hai.

Active recall

Connections

Equations of Motion (constant acceleration) — ye page us ka $a = - g$ special case hai, draw out kiya gaya
Vectors and sign conventions — Step 1 mein ruler
Newton's Second Law — kyun $a = g$ mass cancel ke saath (Step 2)
Projectile Motion — ek steady sideways motion attach karo aur ye vertical half ban jaata hai
Air Resistance and Terminal Velocity — jo straight $v$ – $t$ line ko bend karta hai
Parent: Free fall — g = 9.8 m - s², sign conventions (index 1.1.17)