Free fall — g = 9.8 m - s², sign conventions: Worked examples

1.1.17 · D3 · Physics › Measurement, Vectors & Kinematics › Free fall — g = 9.8 m - s², sign conventions

Kuch bhi shuru karne se pehle, teen tools ko dobara anchor karte hain taaki tumhe kabhi scroll nahi karna pade.

Yahan, $Δ y = y - y_{0}$ ka matlab hai "final position minus starting position" — yeh displacement hai, distance travelled nahi. Yeh distinction matter karti hai aur hum dekhenge yeh kaise problem banati hai.

Scenario matrix

Har free-fall problem actually yeh hai — "main in cells mein se kis mein hoon?" Har cell $v_{0}$ ke sign, landing point ke upar/neeche/shuruat par hone, aur kisi input ke zero hone ya limit pe jaane mein alag hoti hai.

#	Cell class	Kya special hai	Example
1	Pure drop, $v_{0} = 0$	starting velocity zero hai (ek degenerate input)	Ex 1
2	Upar throw kiya, wapas start par land kiya	symmetric flight, $Δ y = 0$	Ex 2
3	Upar throw kiya, start se neeche land kiya	$Δ y < 0$ ; ball apni launch height se neeche jaate hue pass karti hai	Ex 3
4	Neeche throw kiya, dono signs positive	"down positive" convention sab kuch $+$ rakhta hai	Ex 4
5	Same problem, opposite convention	prove karo ki answer choice par depend nahi karta	Ex 5
6	Quadratic with two roots	kaunsa time rakhein? Root ka sign	Ex 6
7	Limiting value	kya hota hai jab $v_{0} \to 0$ ya $g \to$ chhota	Ex 7
8	Real-world word problem	reaction time, ek moving reference	Ex 8
9	Exam twist	do objects, ya ek hidden assumption	Ex 9

Ab hum har cell ko order mein lete hain.

Ex 1 — Cell 1: pure drop ( $v_{0} = 0$ )

Setup (kyun?): "Rest se release" ka matlab hai degenerate input $v_{0} = 0$ — koi throw nahi, gravity sab kaam karti hai. Up positive choose karo, ground ko $y = 0$ par rakho, toh $y_{0} = + 45$ , $a = - 9.8$ .

Time nikalo. Equation (2) use karo $y = 0$ ke saath. $0 = 45 + 0 \cdot t - \frac{1}{2} (9.8) t^{2} .$ Yeh step kyun? Hum us instant ko chahte hain jab ball ground par ho, yaani $y = 0$ . $v_{0} t$ term gayab ho jaata hai kyunki $v_{0} = 0$ hai — notice karo ki degenerate input equation ko simplify karta hai, todta nahi.
$t$ ke liye solve karo. $4.9 t^{2} = 45 ⟹ t^{2} = 9.184 ⟹ t = 3.03 s .$ Yeh step kyun? Simple algebra; sirf positive root physically sense deta hai (negative time release se pehle ka hai).
Impact velocity nikalo. Equation (1) use karo. $v = 0 - 9.8 (3.03) = - 29.7 m/s .$ Yeh step kyun? Sign direction batata hai — negative = neeche hamare convention mein. Speed $29.7 m/s$ hai.

Verify: No-time equation (3) se cross-check karo: $v^{2} = 0 + 2 (9.8) (45) = 882$ , toh $∣ v ∣ = 882 = 29.7 m/s$ . ✓ Match hota hai. Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ . ✓

Ex 2 — Cell 2: upar throw kiya, start par wapas aaya ( $Δ y = 0$ )

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

Setup: Up positive, $y_{0} = 0$ , $v_{0} = + 12$ , $a = - 9.8$ .

Max height. Peak par ball momentarily ruk jaati hai, toh $v = 0$ . Equation (3) use karo: $0 = 1 2^{2} - 2 (9.8) H ⟹ H = \frac{144}{19.6} = 7.35 m .$ $v = 0$ kyun? Figure dekho: velocity shrink hoti hai jab ball upar jaati hai, apex par exactly zero ho jaati hai (plum dot), phir neeche grow karti hai. Turnaround ka instant wohi hai jahan $v = 0$ hai.
Top tak time. Equation (1) se: $0 = 12 - 9.8 t ⟹ t_{up} = 1.22 s$ . Yeh step kyun? Same " $v = 0$ at top" idea, ab time ke liye solve kiya.
Total flight time. Up–down symmetry se (figure mein parabola apex ke baare mein mirror image hai), $t_{total} = 2 t_{up} = 2.45 s$ . Yeh step kyun? Kyunki poore trip mein $Δ y = 0$ hai, equation (2) deta hai $0 = 12 t - 4.9 t^{2} = t (12 - 4.9 t)$ : roots $t = 0$ (launch) aur $t = 2.45$ (return). Do roots woh do times hain jab ball hand height par hoti hai.

Verify: (1) se return speed: $v = 12 - 9.8 (2.45) = - 12 m/s$ — same magnitude, ab neeche ki taraf. ✓ Exactly wohi symmetry jo humne predict ki thi.

Ex 3 — Cell 3: upar throw kiya, start se neeche land kiya ( $Δ y < 0$ )

Setup: Up positive, launch point $y_{0} = 0$ , ground at $y = - 25$ (start se neeche ⟹ negative), $v_{0} = + 8$ , $a = - 9.8$ . Yahan displacement $Δ y = - 25 - 0 = - 25 m$ negative hai — yeh wohi cell hai jahan log mistake karte hain.

Impact speed via (3). $v^{2} = 8^{2} + 2 (- 9.8) (- 25) = 64 + 490 = 554.$ Yeh step kyun? Equation (3) time aur upar-phir-neeche detour ko skip karta hai — perfect jab hum sirf speed chahte hain. Note karo ki do negatives multiply hokar positive bante hain, toh drop speed add karta hai. $∣ v ∣ = 554 = 23.5 m/s (downward, so v = - 23.5) .$
Impact time via (2). $- 25 = 8 t - 4.9 t^{2} ⟹ 4.9 t^{2} - 8 t - 25 = 0.$ Yeh step kyun? Ab hume time chahiye, toh hum position (2) use karte hain. Yeh ek quadratic hai; ise solve karo: $t = \frac{8 \pm 64 + 4 ( 4.9 ) ( 25 )}{2 ( 4.9 )} = \frac{8 \pm 554}{9.8} .$
Root choose karo. $554 = 23.5$ , toh $t = \frac{8 + 23.5}{9.8} = 3.22 s$ ya $t = \frac{8 - 23.5}{9.8} = - 1.58 s$ . Pehla kyun rakhein? Negative time throw se pehle ka hai — physically meaningless yahan. $t = 3.22 s$ rakhte hain.

Verify: $t = 3.22$ ko (1) mein daalo: $v = 8 - 9.8 (3.22) = - 23.6 m/s$ (rounding). ✓ Magnitude step 1 se match karta hai. Aur impact speed ( $23.5$ ) launch speed ( $8$ ) se zyada hai — sensible hai, cliff ne ball ko extra fall distance di.

Ex 4 — Cell 4: neeche throw kiya, all-positive convention

Setup (yeh convention kyun?): Sab kuch neeche move karta hai, toh down positive choose karo — phir $v_{0} = + 6$ , $a = + 9.8$ , aur har number positive rehta hai, sign slips khatam ho jaate hain. Start $y_{0} = 0$ .

Depth via (2). $y = 0 + 6 (2.0) + \frac{1}{2} (9.8) (2.0)^{2} = 12 + 19.6 = 31.6 m .$ Yeh step kyun? Hume time pata hai aur position chahiye — exactly wohi (2) deliver karta hai.
Impact speed via (1). $v = 6 + 9.8 (2.0) = 25.6 m/s .$ Yeh step kyun? Ek known time ke baad velocity — equation (1). Positive ka matlab yahan neeche ki taraf hai (jaisa intended tha).

Verify: (3) se cross-check karo: $v^{2} = 6^{2} + 2 (9.8) (31.6) = 36 + 619.4 = 655.4$ , $655.4 = 25.6 m/s$ . ✓ Depth $31.6 > 19.6$ , toh throw ne depth add ki — sensible.

Ex 5 — Cell 5: SAME problem, up-positive convention

Setup: Up positive. Neeche throw kiya toh $v_{0} = - 6$ . Gravity neeche toh $a = - 9.8$ . Paani start se neeche hai, toh uski position $y = - 31.6$ hogi (negative, kyunki neeche). $y_{0} = 0$ .

$t = 2.0$ par Position via (2). $y = 0 + (- 6) (2.0) + \frac{1}{2} (- 9.8) (2.0)^{2} = - 12 - 19.6 = - 31.6 m .$ Yeh step kyun? Same equation, lekin ab $v_{0}$ aur $a$ dono negative hain. Depth = $∣ - 31.6 ∣ = 31.6 m$ .
$t = 2.0$ par Velocity via (1). $v = - 6 + (- 9.8) (2.0) = - 25.6 m/s .$ Yeh step kyun? Negative kyunki is convention mein yeh neeche ki taraf hai. Speed = $25.6 m/s$ .

Verify: Depth $31.6 m$ aur speed $25.6 m/s$ — Ex 4 se identical. ✓ Physics kabhi is par depend nahi karti ki tum positive kis taraf keh rahe ho; sirf signs flip hote hain. Yahi Vectors and sign conventions discipline ka poora point hai.

Ex 6 — Cell 6: do meaningful roots wala quadratic

Setup: Up positive, $y_{0} = 0$ , $v_{0} = + 20$ , $a = - 9.8$ , target $y = 15$ .

$y = 15$ par equation (2) likho. $15 = 20 t - 4.9 t^{2} ⟹ 4.9 t^{2} - 20 t + 15 = 0.$ Yeh step kyun? Hum height fix karte hain aur poochte hain "kiske liye $t$ ?" — ek quadratic, toh do solutions tak ho sakte hain.
Solve karo. $t = \frac{20 \pm 400 - 4 ( 4.9 ) ( 15 )}{2 ( 4.9 )} = \frac{20 \pm 106}{9.8} .$ $106 = 10.30$ , toh $t = \frac{20 + 10.30}{9.8} = 3.09 s$ ya $t = \frac{20 - 10.30}{9.8} = 0.99 s$ .
DONO roots interpret karo. Figure dekho: ball $15 m$ se ek baar upar jaate hue pass karti hai ( $t = 0.99 s$ , orange dot) aur ek baar neeche aate hue ( $t = 3.09 s$ , teal dot). Dono physical hain — Ex 3 se alag, hum dono rakhte hain. Dono kyun rakhein? Discriminant positive hai aur dono roots launch ke baad positive times hain. Height $15$ apex se neeche hai ( $H = 400/19.6 = 20.4 m$ ), toh ball ise do baar reach karti hai.

Verify: Agar do roots apex ko straddle karte hain, toh unka average time-to-top $t_{up} = 20/9.8 = 2.04 s$ ke barabar hona chahiye. Average $= \frac{0.99 + 3.09}{2} = 2.04 s$ . ✓ Beautiful symmetry check.

Ex 7 — Cell 7: limiting case ( $v_{0} \to 0$ )

Setup: $H = v_{0}^{2} / (2 g)$ use karo $g = 9.8$ ke saath.

Teen speeds plug in karo. $H (10) = \frac{100}{19.6} = 5.10 m, H (1) = \frac{1}{19.6} = 0.051 m, H (0.1) = \frac{0.01}{19.6} = 0.00051 m .$ Yeh step kyun? Value ko shrink hote dekhna dependence ki shape dikhata hai.
Pattern padho. $v_{0}$ ko 10 se divide karo aur $H$ $100$ se girti hai — kyunki $H \propto v_{0}^{2}$ , ek quadratic dependence, linear nahi. Yeh kyun matter karta hai? Yeh humein sensitivity batata hai: zero ke paas, ek chhoti si throw barely ball ko lift karti hai.
Limit lo. $lim_{v_{0} \to 0} H = lim_{v_{0} \to 0} \frac{v _{0}^{2}}{2 g} = 0.$ Yeh step kyun? Yeh "throw up" (Cells 2–3) aur "pure drop" (Cell 1) ke beech ka sanity boundary hai: zero speed ki throw drop hi hai, aur uski rise height exactly $0$ hai. Formulas seamlessly connect hote hain — koi discontinuity nahi.

Verify: Kya Ex 1 ka drop $v_{0} = 0$ se bilkul rise karta hai? Nahi — yeh immediately girna shuru ho jaata hai, $H = 0$ . ✓ Limit se consistent. Do cells exactly $v_{0} = 0$ par milte hain.

Ex 8 — Cell 8: real-world word problem (reaction time)

Setup: Down positive (sirf girna). $v_{0} = 0$ (rest se release), $a = + 9.8$ , $d = 0.18 m$ .

Distance aur time ko link karne ke liye (2) use karo. $d = \frac{1}{2} g t^{2} ⟹ 0.18 = \frac{1}{2} (9.8) t^{2} .$ Yeh step kyun? Hume fallen distance pata hai aur elapsed time chahiye — position equation, $v_{0} = 0$ ke saath (ek word problem mein embedded Cell-1 style degenerate input).
$t$ ke liye solve karo. $t^{2} = \frac{0.36}{9.8} = 0.03673 ⟹ t = 0.192 s .$ Yeh step kyun? Sirf positive root — time negative nahi ho sakta. Reaction time $\approx 0.19 s$ .

Verify: Impact speed agar useful ho: $v = 2 (9.8) (0.18) = 1.88 m/s$ . Time $0.19 s$ human reaction-time band ( $0.15$ – $0.25 s$ ) mein exactly fit hota hai. ✓ Physically believable — yahi literally "ruler drop test" kaise kaam karta hai.

Ex 9 — Cell 9: exam twist (do objects)

Setup: Down positive, $y_{0} = 0$ , ground at $y = + 60$ . Ball A: $v_{0} = 0$ . Ball B: $v_{0} = + 10$ . Dono $a = + 9.8$ .

A ke liye time (ek pure drop, Cell 1). $60 = \frac{1}{2} (9.8) t_{A}^{2} ⟹ t_{A}^{2} = \frac{120}{9.8} = 12.24 ⟹ t_{A} = 3.50 s .$ Yeh step kyun? Koi initial velocity nahi, toh sirf $\frac{1}{2} a t^{2}$ term survive karta hai.
B ke liye time (Cell 4 style). $60 = 10 t_{B} + \frac{1}{2} (9.8) t_{B}^{2} ⟹ 4.9 t_{B}^{2} + 10 t_{B} - 60 = 0.$ $t_{B} = \frac{- 10 + 100 + 4 ( 4.9 ) ( 60 )}{2 ( 4.9 )} = \frac{- 10 + 1276}{9.8} = \frac{- 10 + 35.72}{9.8} = 2.62 s .$ $+$ root kyun rakhein? $-$ root negative time deta hai (launch se pehle), discard karo.
Difference. $Δ t = t_{A} - t_{B} = 3.50 - 2.62 = 0.88 s$ . Yeh step kyun? Question impacts ke beech gap ke baare mein poochh raha hai.

Verify: Sanity: B ke paas speed ka head start hai, toh use pehle land karna chahiye ( $t_{B} < t_{A}$ ) — ✓. A ki impact speed check karo $v_{A} = 2 (9.8) (60) = 34.3 m/s$ vs B ki $v_{B} = 1 0^{2} + 2 (9.8) (60) = 1276 = 35.7 m/s$ ; B faster hai, pehle land karne ke consistent. ✓

Recap: kaun sa cell kaun sa tha

Recall Map each example back to its matrix cell

Ex 1 → Cell 1 (drop, $v_{0} = 0$ ) · Ex 2 → Cell 2 ( $Δ y = 0$ symmetric) · Ex 3 → Cell 3 ( $Δ y < 0$ ) · Ex 4 → Cell 4 (down positive) · Ex 5 → Cell 5 (convention swap) · Ex 6 → Cell 6 (two roots) · Ex 7 → Cell 7 (limit $v_{0} \to 0$ ) · Ex 8 → Cell 8 (word problem) · Ex 9 → Cell 9 (two objects).

Active recall

Kaun sa equation time ko puri tarah skip karta hai?

v^{2} = v_{0}^{2} + 2 a Δ y

(equation 3).

Ex 3 mein, impact speed launch speed se zyada kyun thi?

Cliff ne launch point se neeche extra fall distance add ki (

Δ y < 0

), toh

2 a Δ y

positive hai aur

v_{0}^{2}

mein add hota hai.

Position quadratic ke dono roots kab rakhein?

Jab dono positive times hon aur object physically us height se do baar pass kare (target apex se neeche ho).

Jab

v_{0} \to 0

, max height

H = v_{0}^{2} /2 g

kya approach karta hai, aur yeh kyun sense deta hai?

H \to 0

; zero-speed "throw" sirf ek drop hi hai, jo bilkul rise nahi karta.

Final answer up-positive vs down-positive choose karne par depend karta hai kya?

Nahi — Ex 4 aur Ex 5 identical depth aur speed dete hain; sirf intermediate signs flip hote hain.

Connections

Parent: Free fall & sign conventions
Equations of Motion (constant acceleration) — har example par use kiye gaye teen tools
Vectors and sign conventions — kyun Ex 5 kaam karta hai
Projectile Motion — yeh vertical cases $y$ -component ban jaate hain
Newton's Second Law — $a = g$ ka origin
Air Resistance and Terminal Velocity — jab yeh idealized answers break hote hain

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