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

False (vacuum mein). Kyunki $ma=mg$ se $a=g$ milta hai, mass cancel ho jaata hai aur har object identically accelerate karta hai; "heavy cheez tez girti hai" wala effect jo tumhe hawa mein dikhta hai woh air resistance hai, gravity nahi.

False. Free fall ka matlab hai gravity hi ek force hai — yeh upar jaate waqt, top par, aur neeche aate waqt — teeno mein sach hai. Object decelerate kar raha hota hai, lekin sirf gravity hi act kar rahi hoti hai.

False. Sirf velocity momentarily zero hoti hai; acceleration abhi bhi $-g$ hai. Agar $a$ zero hota toh ball wahan hamesha ke liye hover karti instead of neeche aane ke.

False. Gravity neeche point karti hai, isliye upar positive lene par acceleration $a=-g=-9.8{\text{m/s}}^2$ hota hai. Number $9.8$ ek magnitude hai; sign tumhare chosen axis se decide hota hai.

True (bina hawa ke). Time-free relation $v^2=v_0^2-2g\Delta y$ mein $\Delta y=0$ rakhne par, waapsi speed ka magnitude $v_0$ ke barabar hota hai — lekin velocity ab $-v_0$ hai (neeche ki taraf directed).

False. Yeh sirf relabel karta hai ki hum kaunsi direction ko $+$ kehte hain; actual motion identical hai. Down positive ke saath hum simply $a=+g$ likhte hain aur downward velocities ko positive maante hain — dekho Vectors and sign conventions.

True. Vertical aur horizontal motion independent hoti hain; vertical drop $y=y_0-\frac{1}{2}g{t}^2$ ko follow karta hai, horizontal speed se fark nahi padta. Yahi Projectile Motion ka core idea hai.

False. Upar jaate waqt velocity positive hoti hai (upar move kar rahi hai); top par zero se guzarti hai aur tabhi negative hoti hai. Acceleration poore time negative rehta hai, velocity nahi.

Equation sirf yeh kehti hai ki velocity us instant $t=v_0/g$ par instantaneously zero hai; ek moment baad $t$ bada ho jaata hai toh $v$ negative ho jaati hai. Zero velocity ek passing instant hai, resting state nahi.

$9.8$ metres-per-second per second hai, per metre nahi. Yeh har second mein $9.8\text{m/s}$ speed gain karta hai, har metre distance mein nahi.

Acceleration ki direction aur velocity ki direction independent hoti hain. Upar jaate waqt $a=-g$ hai jabki velocity upward (positive) hai; negative acceleration hi upward motion ko slow karti hai.

Signs chosen axis se match karne chahiye. Agar upar positive hai, toh neeche ki taraf phenkne ka matlab hai $v_0=-5$; $+5$ likhna conventions ko mix karta hai aur baad ke saare results kharab kar deta hai.

Ek $\frac{1}{2}$ ka factor. $0=v_0^2-2gH$ se $H=v_0^2/(2g)$ milta hai; constant acceleration ko integrate karne wala "2" drop ho gaya.

Free fall forces se define hota hai, motion ki direction se nahi. Engines band aur hawa nahi matlab sirf gravity act kar rahi hai, isliye coasting rocket free fall mein HAI, chahe woh upar ja raha ho.

Same acceleration, same impact speed nahi. Impact speed drop height par depend karti hai $v^2=2g\Delta y$ ke through; 20 m wali ball tez pahunchti hai.

Gravity ki pull mass ke saath badhti hai ($F=mg$), lekin accelerate hone ki resistance bhi ($F=ma$). $ma=mg$ set karne par $m$ cancel ho jaata hai, $a=g$ bachta hai — dono mass effects exactly offset karte hain. Dekho Newton's Second Law.

Motion symmetric hai: wahi constant acceleration $-g$ jo upar jaate waqt speed $v_0$ remove karta hai, neeche aate waqt exactly $v_0$ restore karta hai same height par, isliye dono legs $t=v_0/g$ lete hain.

Kyunki velocity, displacement, aur acceleration directional (vectors) hote hain, aur sign hi 1-D mein direction encode karne ka tarika hai. Fixed convention ke bina "upar" aur "neeche" mix ho jaate hain aur equations nonsense deti hain — Vectors and sign conventions.

Kyunki Earth ki surface ke paas $g$ essentially constant hai, toh $a$ constant hai, jo exactly woh condition hai jiske under un kinematic equations derive ki gayi hain.

Earth par air resistance pankh ko uske tiny weight ke relative strongly oppose karti hai; airless Moon par woh force nahi hai, toh dono sirf gravity experience karte hain aur identically girte hain. Yahan Air Resistance and Terminal Velocity ideal ko break karta hai.

Acceleration measure karta hai ki velocity kaise change ho rahi hai, uski current value nahi. Jis instant $v=0$ hoti hai, velocity $+$ se $-$ ki taraf ja rahi hoti hai, jo ek downward (negative) change hai — isliye $a=-g$.

Extra initial speed $v_0$, final speed mein $v^2=v_0^2+2g\Delta y$ ke through add hoti hai, lekin acceleration $g$ hi rehta hai kyunki gravity unchanged hai — tumne starting condition badla, force nahi.

Pehle se hi $g$ downward. Chahe $t=0$ par speed zero ho, gravity immediately act karti hai, isliye acceleration $g$ pehle instant se hi hota hai — yeh kabhi "ramp up" nahi karta.

Nahi. Velocity sirf top par zero hoti hai, lekin wahan acceleration $-g$ hai; ideal free fall mein koi aisa moment nahi hota jab acceleration vanish ho jaaye.

$v^2=v_0^2$, toh $v=\pm v_0$. Physically meaningful root $-v_0$ hai (downward), jo confirm karta hai ki ball launch speed ke saath lekin opposite direction mein waapas aati hai.

$v_0=\sqrt{2gH}$ — same speed jo ek dropped ball us height par gain karti hai, kyunki up-trip aur $H$ par ek drop mirror images hain.

Haan, contact tak sirf gravity hi force hai, isliye $a=g$. Sudden change impact par hoti hai jab ground normal force exert karta hai, free-fall phase khatam karta hai.

Yeh fail ho jaata hai: air resistance speed ke saath badhti hai jab tak gravity ko balance nahi karti, zero net acceleration aur ek constant terminal velocity deti hai. Free fall assume karta hai ki yeh force absent hai.

Negative. Down positive ke saath, upward motion negative direction hai, isliye upward throw $v_0<0$ se shuru hoti hai jabki $a=+g$ hota hai; physics up-positive ke saath relabeling ke baad match karti hai.