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

1.1.17 · D4 · Physics › Measurement, Vectors & Kinematics › Free fall — g = 9.8 m - s², sign conventions

Intuition Is page ko kaise use karein

Neeche har problem ek clean scenario batata hai. Pehle khud try karo, phir collapsible Solution callout kholke apna poora working check karo. Problems difficulty mein badhte jaate hain: L1 = idea pehchano → L5 = sab kuch combine karo. Poore time, jab tak alag se na bola jaaye, $g = 9.8 m/s^{2}$ use karo, air resistance ignore karo (woh free fall tod deta — dekho Air Resistance and Terminal Velocity), aur hamesha line one par apna sign convention declare karo — bilkul waise jaise humne Vectors and sign conventions mein set up kiya tha. Jinhe hum use karte hain woh teen equations seedha Equations of Motion (constant acceleration) se aate hain: $v = v_{0} + a t, y = y_{0} + v_{0} t + \frac{1}{2} a t^{2}, v^{2} = v_{0}^{2} + 2 a Δ y .$ Yahan $a = \pm g$ is baat par depend karta hai ki tumne kaunsa direction positive call kiya.

Shuru karne se pehle, ek reminder ki har symbol ka matlab kya hai, taaki kuch bhi undefined na reh jaaye:

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

Upar wali figure har problem ke peeche ki mental picture hai: ek vertical line par ek object, ek arrow jo woh positive direction dikhata hai jo tumne choose ki, aur gravity ke liye ek fixed downward arrow. Baaki sab kuch bas teen boxed equations mein numbers plug karna hai.

Level 1 — Recognition

L1.1 — $a$ ka sign kya hai?

Ek rock ko ek bridge se drop kiya jaata hai. Tum neeche ko positive direction call karne ka decide karte ho. Acceleration $a$ (uske sign ke saath) kya hai?

Recall Solution

Setup: Down positive hai. Gravity neeche point karti hai, jo ab positive direction hai. Isliye $a = + g = + 9.8 m/s^{2}$ . Kyun: acceleration hamesha physically Earth ki taraf point karta hai; sign sirf yeh record karta hai ki kya yeh tumhare chosen axis se agree karta hai ya nahi. Down positive ⟹ dono agree karte hain ⟹ plus.

L1.2 — Velocity equation read karna

Up positive use karte hue, ek ball ko $v_{0} = + 12 m/s$ se upar throw kiya jaata hai. Time $t$ par uski velocity $v$ ke liye equation likho, aur $t = 0$ par velocity batao.

Recall Solution

Setup: Up positive ⟹ $a = - g = - 9.8$ . $v = v_{0} + a t$ mein plug karo: $v = 12 - 9.8 t .$ $t = 0$ par: $v = 12 - 0 = + 12 m/s$ — bilkul launch speed, upar point karta hua. Achha sanity check hai.

L1.3 — Bilkul top par kya sach hai?

Seedhi upar throw ki gayi ball apne highest point par pahunchti hai. Fill karo: us instant par, velocity ____ hai aur acceleration ____ hai.

Recall Solution

Velocity $v = 0$ (ball momentarily vertically chalna band kar deti hai). Acceleration $a = - g = - 9.8 m/s^{2}$ — gravity kabhi switch off nahi hoti. $v = 0$ kyun lekin $a \neq = 0$ : velocity zero hona matlab hai "abhi right now move nahi kar rahi"; acceleration nonzero hona matlab hai "phir bhi velocity change ho rahi hai", yahi wajah hai ball neeche girna shuru karti hai.

Level 2 — Application

L2.1 — Ek height se drop kiya

Ek stone ko $45 m$ ki height se rest se drop kiya jaata hai. Ground se takraane mein kitna time lagta hai?

Recall Solution

Setup: Up positive, $y_{0} = 45$ , ground $y = 0$ par, $v_{0} = 0$ ("dropped"), $a = - 9.8$ . $y = y_{0} + v_{0} t + \frac{1}{2} a t^{2}$ use karo: $0 = 45 + 0 - \frac{1}{2} (9.8) t^{2} ⟹ 4.9 t^{2} = 45 ⟹ t^{2} = 9.184.$ $t = 9.184 = 3.03 s .$ Hum positive root rakhte hain kyunki time forward chalta hai.

L2.2 — Impact speed

Usi $45 m$ drop ke liye, ground se takraane par speed find karo.

Recall Solution

Sabse fast route: $v^{2} = v_{0}^{2} + 2 a Δ y$ . Yahan $Δ y = y - y_{0} = 0 - 45 = - 45$ (woh neeche move hua, hamare positive axis ke opposite), $v_{0} = 0$ , $a = - 9.8$ : $v^{2} = 0 + 2 (- 9.8) (- 45) = 882 ⟹ v = - 882 = - 29.7 m/s .$ Negative root kyun: stone impact par neeche ja raha hai, hamare positive (up) direction ke opposite. Uski speed (magnitude) $29.7 m/s$ hai.

L2.3 — Neeche throw kiya

Ek ball ko $60 m$ ki cliff se $8 m/s$ se neeche throw kiya jaata hai. Uski impact speed find karo.

Recall Solution

Setup — down positive (sab kuch ek taraf girta hai, isliye sare numbers positive rehte hain → kam slips): $a = + 9.8$ , $v_{0} = + 8$ , $Δ y = + 60$ . $v^{2} = v_{0}^{2} + 2 a Δ y = 8^{2} + 2 (9.8) (60) = 64 + 1176 = 1240.$ $v = 1240 = 35.2 m/s (downward) .$

Level 3 — Analysis

L3.1 — Max height aur total flight time

Ek ball ko ground level se $v_{0} = 20 m/s$ se seedha upar throw kiya jaata hai. (a) Maximum height find karo. (b) Jab tak woh ground par wapas aaye tab tak ka total time find karo.

Recall Solution

Setup: Up positive, $y_{0} = 0$ , $v_{0} = + 20$ , $a = - 9.8$ . (a) Max height. $H$ maximum height ho — yaani launch point se top tak displacement, isliye $H = Δ y = y_{top} - y_{0}$ . Kyun hum ise time ke bina find kar sakte hain: top par $v = 0$ (ball momentarily ruk jaati hai), aur time-free equation $v^{2} = v_{0}^{2} + 2 a Δ y$ velocity ko directly displacement se link karta hai. $v = 0$ aur $Δ y = H$ set karo: $0 = 2 0^{2} + 2 (- 9.8) H ⟹ H = \frac{400}{19.6} = 20.4 m .$ (b) Total time: path symmetric hai — upar jaane ka time neeche aane ke time ke barabar hai. $v = v_{0} + a t$ se top tak time: $0 = 20 - 9.8 t_{up} ⟹ t_{up} = \frac{20}{9.8} = 2.04 s .$ $t_{total} = 2 t_{up} = 4.08 s .$

Neeche wala plot yeh poori flight dikhata hai. Blue curve visually trace karo: curve utha hua hai, yellow dot par flat ho jaata hai (jahan tangent horizontal hai — woh $v = 0$ hai), phir mirror-symmetrically wapas girta hai. Dashed vertical line $t_{up} = 2.04 s$ par parabola ko do equal halves mein split karti hai — yeh visual symmetry exactly wahi hai kyun upar jaane ka time neeche aane ke time ke barabar hai, aur kyun peak flight ke dead centre par baithta hai.

L3.2 — Ek given height par velocity

Usi throw ke liye ( $v_{0} = 20$ ), ball ki velocity kya hai jab woh height $y = 15 m$ par hai, upar jaate waqt? Neeche aate waqt usi height par kya?

Recall Solution

$v^{2} = v_{0}^{2} + 2 a Δ y$ use karo jisme $Δ y = 15 - 0 = 15$ : $v^{2} = 400 + 2 (- 9.8) (15) = 400 - 294 = 106 ⟹ v = \pm 10.3 m/s .$ Equation do signs deta hai kyunki ball $15 m$ se do baar guzarti hai:

Upar jaate waqt: $v = + 10.3 m/s$ (upar ja rahi hai).
Neeche aate waqt: $v = - 10.3 m/s$ (same speed, neeche ja rahi hai). Yeh wahi up–down symmetry hai, algebra mein se automatically pop out ho rahi hai. Upar wali figure mein, $y = 15 m$ par ek horizontal line parabola ko do jagah cut karegi — ek rising half par, ek falling half par — yahi two-crossing fact graphically dikh raha hai.

Level 4 — Synthesis

L4.1 — Do balls, staggered release

Ball A ko $50 m$ tower se rest se drop kiya jaata hai. Ek second baad, ball B ko usi point se drop kiya jaata hai. Jab A $3 s$ ke liye gir chuka hai, do balls kitni door hain?

Recall Solution

Setup — down positive, origin release point par. Distance fallen $= \frac{1}{2} g t^{2}$ jahan $t$ hai har ball ka apna fall time.

Ball A $t_{A} = 3 s$ se gir rahi hai: $y_{A} = \frac{1}{2} (9.8) (3)^{2} = 4.9 \times 9 = 44.1 m$ .
Ball B $1 s$ baad shuru hui, isliye is instant par $t_{B} = 3 - 1 = 2 s$ : $y_{B} = \frac{1}{2} (9.8) (2)^{2} = 4.9 \times 4 = 19.6 m$ .
Separation $= y_{A} - y_{B} = 44.1 - 19.6 = 24.5 m$ . (Check karo ki A abhi bhi airborne hai: woh ground hit karta hai jab $\frac{1}{2} g t^{2} = 50 \Rightarrow t = 3.19 s > 3$ . Good.) Insight: do objects ke beech gap jo ek fixed time apart drop hue, badhta rehta hai, kyunki earlier ball hamesha faster move kar rahi hoti hai — time mein head start distance mein ek ever-larger head start ban jaata hai.

L4.2 — Mid-air mein milna

Ground se, ek ball $v_{0} = 25 m/s$ se upar throw ki jaati hai. Usi instant par, ek doosri ball $40 m$ ki height se directly upar rest se drop ki jaati hai. Kis time par woh milte hain, aur kis height par?

Recall Solution

Setup: Up positive, common origin ground par, dono ke liye $t = 0$ .

Thrown ball: $y_{1} = 0 + 25 t - \frac{1}{2} (9.8) t^{2}$ .
Dropped ball: $y_{2} = 40 + 0 - \frac{1}{2} (9.8) t^{2}$ . Woh milte hain jab $y_{1} = y_{2}$ . $- \frac{1}{2} (9.8) t^{2}$ terms identical hain, isliye woh cancel ho jaate hain: $25 t = 40 ⟹ t = \frac{40}{25} = 1.6 s .$ Kyun woh gravity terms cancel hote hain — physics, sirf algebra nahi: dono balls exactly same acceleration $- g$ feel karte hain (mass kabhi enter nahi hua — dekho Newton's Second Law). Isliye gravity dono ko har instant par identical amount $\frac{1}{2} g t^{2}$ se neeche khichti hai. Agar tum dono ke beech gap dekho, woh shared falling motion kuch contribute nahi karta — yeh dono balls ko equally shift karta hai. Gap isliye purely throw speed $25 m/s$ par band hota hai, bilkul jaise relative motion ke liye gravity hai hi nahi. $40 m$ door se shuru karke aur $25 m/s$ par close karte hue, woh $40/25 = 1.6 s$ ke baad milte hain — jo upar wali equation hai, ab uske peeche ek picture hai. Height: $y = 40 - \frac{1}{2} (9.8) (1.6)^{2} = 40 - 4.9 \times 2.56 = 40 - 12.5 = 27.5 m$ .

Level 5 — Mastery

L5.1 — Moving platform se upar throw ki gayi ball… phir edge se past drop hui

Ek ball ko ek cliff ke edge se upar $v_{0} = 15 m/s$ se throw kiya jaata hai jo neeche ground se $30 m$ upar hai. Find karo: (a) launch se lekar ground par land karne tak ka total time, (b) landing se theek pehle uski velocity.

Recall Solution

Setup: Up positive, origin launch point (cliff edge) par, isliye ground $y = - 30$ par hai. $v_{0} = + 15$ , $a = - 9.8$ . (a) Hume woh moment chahiye jab ball ground tak pahunche, yaani $y = - 30$ . $y_{0} = 0$ ke saath $y = y_{0} + v_{0} t + \frac{1}{2} a t^{2}$ mein substitute karo: $- 30 = 15 t - \frac{1}{2} (9.8) t^{2} ⟹ - 30 = 15 t - 4.9 t^{2} .$ Kyun $4.9 t^{2} - 15 t - 30 = 0$ mein rearrange karein: yeh $t$ mein ek quadratic hai, aur quadratic formula tabhi kaam karta hai jab har term ek taraf standard shape $A t^{2} + B t + C = 0$ mein ho. Sare terms left side par move karo (jo unke signs flip karta hai) taaki $A = 4.9$ , $B = - 15$ , $C = - 30$ mile. Solve karo $t = \frac{- B \pm B ^{2} - 4 A C}{2 A}$ se: $t = \frac{15 \pm 1 5 ^{2} + 4 ( 4.9 ) ( 30 )}{2 ( 4.9 )} = \frac{15 \pm 225 + 588}{9.8} = \frac{15 \pm 813}{9.8} = \frac{15 \pm 28.5}{9.8} .$ Do roots hain $t = \frac{43.5}{9.8} = 4.44 s$ aur $t = \frac{- 13.5}{9.8} = - 1.38 s$ . Kyun hum $+ 4.44 s$ root rakhte hain aur $- 1.38 s$ discard karte hain: equation time mein ball ki height ko ek smooth parabola ki tarah describe karta hai. Yeh ground level $y = - 30$ ko do instants par cross karta hai — lekin unme se ek ( $t = - 1.38 s$ ) uss waqt ka hai jab humne ball release bhi nahi ki thi, ek imaginary "past" jahan parabola, agar peeche chalaayi jaaye, toh bhi us height se guzri hoti. Physically motion $t = 0$ par shuru hui, isliye sirf future crossing real hai. Isliye $t = 4.44 s$ . (b) Landing par velocity $v = v_{0} + a t$ se: $v = 15 - 9.8 (4.44) = 15 - 43.5 = - 28.5 m/s .$ Negative ⟹ neeche ja raha hai; impact speed $= 28.5 m/s$ . Time-free equation se cross-check: $v^{2} = 1 5^{2} + 2 (- 9.8) (- 30) = 225 + 588 = 813$ , isliye $∣ v ∣ = 813 = 28.5 m/s$ ✓ — bilkul agree karta hai.

Neeche wali picture "broken symmetry" ko concrete banati hai. Blue curve trace karo: woh yellow dot tak utha hua hai (top, jahan tangent flat hai aur $v = 0$ ), launch height (dashed white line) se wapas guzar kar girta hai, aur phir iske neeche keep going karta hai, green ground line tak jo $30 m$ neeche hai. Dekho kaise falling stretch visibly longer hai rising stretch se — woh extra length ball ka uske starting point ke neeche ka safar hai, aur exactly yahi reason hai kyun simple "twice the time to the top" rule yahan fail karta hai.

L5.2 — Jahan naive "up = down time" symmetry toot jaati hai

L5.1 cliff throw ke liye, ek student claim karta hai "upar jaane ka time = neeche aane ka time, isliye flight symmetric hai." Quantitatively explain karo kyun total time simply time to the top ka do guna nahi hai, aur dono numbers do.

Recall Solution

Top tak time (jahan $v = 0$ ): $t_{up} = v_{0} / g = 15/9.8 = 1.53 s$ . Agar ball launch height par wapas aati, toh neeche aane mein bhi $1.53 s$ lagta, dete $3.06 s$ . Lekin ground launch point se $30 m$ neeche hai — ball apne starting level se aage girti rehti hai. Isliye descent longer hai: $t_{down} = t_{total} - t_{up} = 4.44 - 1.53 = 2.91 s .$ Clean up–down symmetry sirf equal heights ke beech hold karti hai (figure par phir se dekho: mirror symmetry poori tarah dashed launch line ke upar rehti hai). Yahan start aur finish heights $30 m$ se differ karti hain, isliye descent leg bada hai. Symmetry ek special case hai, koi law nahi.

Connections

Equations of Motion (constant acceleration) — teen equations jo yahan har solution power karte hain.
Vectors and sign conventions — har problem ki line one: ek positive direction choose karo.
Projectile Motion — yeh vertical calculations kisi bhi projectile ka $y$ -component hain.
Newton's Second Law — woh reason ki $a = g$ har mass ke liye.
Air Resistance and Terminal Velocity — "ignore air" assumption hatao aur yeh numbers change ho jaate hain.

Recall Self-test checklist

Kya tumne, har problem ke liye: (1) pehle up-vs-down positive state kiya, (2) $a$ ko correct sign diya, (3) $Δ y$ ka sign us axis se match kiya, (4) kisi bhi square root ya quadratic ka physically correct root rakha? Answer ::: Agar charon ke liye haan, toh sign errors — free fall mein marks lose karne ka #1 source — tumhare peeche hain.