Projectile motion — horizontal - vertical independence, full derivation: Exercises

1.1.19 · D4 · Physics › Measurement, Vectors & Kinematics › Projectile motion — horizontal - vertical independence, full

Yeh page ek self-test ladder hai. Har rung pichle se mushkil hai. Har problem ko solution chhupaake try karo, phir [!recall]- callout kholke har step check karo. Saare symbols parent note the main derivation se aaye hain — kuch naya assume nahi kiya gaya.

Quick symbol reminder taaki pehli line samajh aaye:

Level 1 — Recognition

Kya tum sahi formula pick karke plug in kar sakte ho?

L1·Q1 — Components padho

Ek ball $u = 50 m/s$ , $θ = 3 7^{\circ}$ par ground se nikalta hai ( $cos 3 7^{\circ} = 0.8$ , $sin 3 7^{\circ} = 0.6$ use karo). $u_{x}$ aur $u_{y}$ likho.

Neeche ki picture poore problem ki object hai: launch velocity ek arrow hai, aur uske do shadows woh pieces hain jo hum compute karte hain.

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

Recall Solution

WHAT: launch arrow ko uske do shadows mein split karo. WHY: do axes independently evolve hoti hain, isliye har piece alag chahiye. $u_{x} = u cos θ = 50 \times 0.8 = 40 m/s$ $u_{y} = u sin θ = 50 \times 0.6 = 30 m/s$ Figure s01 upar dekho: burnt-orange arrow $u$ hai, teal arrow $u_{x}$ hai (floor-shadow), plum arrow $u_{y}$ hai (wall-shadow). Woh jo right triangle banaate hain woh exactly wahi hai jo $cos$ (adjacent/hypotenuse) aur $sin$ (opposite/hypotenuse) measure karta hai.

L1·Q2 — Time of flight

Same ball ( $u_{y} = 30 m/s$ ). Level ground. $T$ nikalo.

Recall Solution

WHY sirf $u_{y}$ ? Sirf vertical motion decide karta hai kab $y$ wapas $0$ par aata hai; horizontal glide timing se irrelevant hai. $T = \frac{2 u s i n θ}{g} = \frac{2 \times 30}{10} = 6 s$

L1·Q3 — Max height aur range

Same ball. $H$ aur $R$ nikalo.

Recall Solution

WHY $H = u_{y}^{2} / (2 g)$ ? Formula $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ mein numerator $(u sin θ)^{2}$ hai. Lekin $u sin θ$ hi vertical shadow $u_{y}$ hai jo humne define kiya tha. Toh $u^{2} sin^{2} θ = u_{y}^{2}$ , aur hum likh sakte hain $H = \frac{u _{y}^{2}}{2 g}$ — woh number $u_{y} = 30$ use karke jo already hamare paas hai. $H = \frac{u ^{2} s i n ^{2} θ}{2 g} = \frac{u _{y}^{2}}{2 g} = \frac{3 0 ^{2}}{2 \times 10} = \frac{900}{20} = 45 m$ $R = u_{x} \cdot T = 40 \times 6 = 240 m$ (Range formula se check karo: $R = \frac{u ^{2} sin 2 θ}{g} = \frac{2500 \times sin 7 4 ^{\circ}}{10} = \frac{2500 \times 0.9613}{10} \approx 240 m$ ✓.)

Level 2 — Application

Do ya teen steps chain karo.

L2·Q1 — Kisi given instant par velocity

Ek projectile $u = 20 m/s$ , $θ = 6 0^{\circ}$ par launch kiya gaya hai ( $sin 6 0^{\circ} = 0.866$ , $cos 6 0^{\circ} = 0.5$ ). $t = 1 s$ par velocity ki speed aur direction nikalo.

Recall Solution

WHAT: $t = 1$ par har velocity component nikalo, phir recombine karo. $u_{x} = 20 \times 0.5 = 10 m/s$ (constant — koi horizontal force nahi). $u_{y} = 20 \times 0.866 = 17.32 m/s$ . $t = 1 s$ par: $v_{x} = u_{x} = 10 m/s, v_{y} = u_{y} - g t = 17.32 - 10 \times 1 = 7.32 m/s$ Speed = arrow ka magnitude (do shadows par Pythagoras): $v = v_{x}^{2} + v_{y}^{2} = 1 0^{2} + 7.3 2^{2} = 100 + 53.6 = 153.6 = 12.39 m/s$ Direction horizontal se upar (woh angle jiska tan rise/run ho): $ϕ = arctan (\frac{v _{y}}{v _{x}}) = arctan (\frac{7.32}{10}) = 36. 2^{\circ}$ Abhi bhi chadh raha hai (kyunki $v_{y} > 0$ ), bas launch se kam steep hai.

L2·Q2 — Cliff se horizontal launch

Ek stone $80 m$ uchi cliff se horizontally $u = 15 m/s$ par throw kiya gaya hai. Flight time aur base se kitni door yeh land karta hai, nikalo.

Neeche ki figure dikhati hai kyun timing ek pure free-fall problem hai: sideways glide hataao aur vertical drop ek stone release from rest ke identical hai.

Recall Solution

WHY $θ = 0$ ? "Horizontally" matlab launch arrow flat hai: $u_{x} = 15$ , $u_{y} = 0$ . Sign convention choose karna. Master equation mein, upar positive hai aur drop $y = u_{y} t - \frac{1}{2} g t^{2}$ likha hai; $80 m$ neeche land karna matlab $y = - 80$ hai, jo $- 80 = - 5 t^{2}$ deta hai. Yahan yeh cleaner hai ki distance downward ko positive maano: tab fallen distance $d$ is tarah badhti hai $d = \frac{1}{2} g t^{2}$ (minus sign plus ban jaata hai kyunki "neeche" ab positive direction hai, aur gravity usi taraf point karta hai). Dono conventions same $t$ dete hain; hum "down positive" sirf minus sign carry karne se bachne ke liye lete hain. Vertically rest se start karke ( $u_{y} = 0$ ): $80 = \frac{1}{2} g t^{2} = \frac{1}{2} (10) t^{2} = 5 t^{2} \Rightarrow t^{2} = 16 \Rightarrow t = 4 s$ Horizontal: constant speed × time $x = u_{x} t = 15 \times 4 = 60 m$ Figure s02 upar arc dikhata hai: vertical drop us free-fall dashed line se identical hai jo same instant par drop ki gayi — woh ground par saath pahunchte hain.

L2·Q3 — Same range, do angles

Dikhao ki $θ = 2 0^{\circ}$ aur $θ = 7 0^{\circ}$ same speed par same range dete hain, aur $u = 30 m/s$ ke liye check karo.

Recall Solution

WHY match karte hain: $R = \frac{u ^{2} sin 2 θ}{g}$ . $2 0^{\circ}$ ke liye: $sin 4 0^{\circ}$ . $7 0^{\circ}$ ke liye: $sin 14 0^{\circ} = sin (18 0^{\circ} - 4 0^{\circ}) = sin 4 0^{\circ}$ . Same value ⇒ same range. $R = \frac{3 0 ^{2} \times s i n 4 0 ^{\circ}}{10} = \frac{900 \times 0.6428}{10} = 57.85 m (dono angles)$ Lekin unka time differ karta hai: $sin 7 0^{\circ} > sin 2 0^{\circ}$ , isliye $7 0^{\circ}$ wali ball zyada der upar rehti hai aur zyada high fly karti hai.

Level 3 — Analysis

Logic ulta karo, ya independence ko ek twist ke saath combine karo.

L3·Q1 — Flight data se launch angle nikalo

$u = 40 m/s$ par launch kiya gaya ek projectile level ground par time of flight $T = 4 s$ hai. $θ$ aur range nikalo.

Recall Solution

WHAT: angle solve karne ke liye time-of-flight formula ko invert karo. $T = \frac{2 u s i n θ}{g} \Rightarrow sin θ = \frac{g T}{2 u} = \frac{10 \times 4}{2 \times 40} = \frac{40}{80} = 0.5$ WHY $arcsin$ ? Yeh jawaab deta hai "kis angle ka yeh sine hai?" — woh inverse jo $sin$ ko undo karta hai. $θ = arcsin (0.5) = 3 0^{\circ}$ Range: $R = u cos θ \cdot T = 40 cos 3 0^{\circ} \times 4 = 40 \times 0.866 \times 4 = 138.6 m$

L3·Q2 — Do stones, ground par race

$45 m$ cliff se, stone A drop kiya gaya aur stone B horizontally $12 m/s$ par throw kiya gaya, dono same instant par. Kaun pehle land karta hai, aur unke landing points ke beech horizontal gap kya hai?

Recall Solution

Key insight (independence): B ka horizontal throw uski vertical motion ko nahi touch karta. Dono stones identical vertical equation $45 = \frac{1}{2} g t^{2}$ share karte hain. $45 = 5 t^{2} \Rightarrow t^{2} = 9 \Rightarrow t = 3 s DONO ke liye$ Toh woh simultaneously land karte hain — koi nahi jeet ta. Stone A cliff base par land karta hai ( $x_{A} = 0$ ). Stone B $x_{B} = 12 \times 3 = 36 m$ travel karta hai. Gap: $Δ x = 36 - 0 = 36 m$

L3·Q3 — Velocity known hone par height

Ek ball $u = 25 m/s$ , $θ = 5 3^{\circ}$ par launch hui ( $sin 5 3^{\circ} = 0.8$ , $cos 5 3^{\circ} = 0.6$ ). Way up par uski speed $20 m/s$ kis height par hai?

Recall Solution

WHAT: $v_{x}$ fixed hai, isliye speed condition $v_{y}$ pin karta hai; phir us $v_{y}$ par height nikalo. $u_{x} = 25 \times 0.6 = 15 m/s$ (constant), $u_{y} = 25 \times 0.8 = 20 m/s$ . Speed $v = 20$ : $v_{x}^{2} + v_{y}^{2} = 2 0^{2} \Rightarrow 1 5^{2} + v_{y}^{2} = 400 \Rightarrow v_{y}^{2} = 400 - 225 = 175$ . Height vertical energy relation se $v_{y}^{2} = u_{y}^{2} - 2 g y$ (1-D kinematics se): $y = \frac{u _{y}^{2} - v _{y}^{2}}{2 g} = \frac{2 0 ^{2} - 175}{2 \times 10} = \frac{400 - 175}{20} = \frac{225}{20} = 11.25 m$

Level 4 — Synthesis

Parametric equations se directly answer banao.

L4·Q1 — Cliff se angle par launch

$60 m$ tower ke top se ek projectile $u = 25 m/s$ , $θ = 3 7^{\circ}$ par fire kiya gaya ( $sin 3 7^{\circ} = 0.6, cos 3 7^{\circ} = 0.8$ ). Total flight time aur tower base se horizontal distance nikalo jahan yeh land karta hai.

Neeche ki figure poora arc dikhata hai jo launch line se neeche girti hai — isliye hum level-ground shortcut use nahi kar sakte.

Recall Solution

WHAT & WHY: landing launch se $60 m$ neeche hai, isliye hum parametric equation mein $y = - 60$ set karte hain (upar positive, origin launch point par). $u_{x} = 25 \times 0.8 = 20 m/s$ , $u_{y} = 25 \times 0.6 = 15 m/s$ . $y = u_{y} t - \frac{1}{2} g t^{2} \Rightarrow - 60 = 15 t - 5 t^{2}$ Standard quadratic mein rearrange karo: $5 t^{2} - 15 t - 60 = 0 \Rightarrow t^{2} - 3 t - 12 = 0$ WHY quadratic formula? Do unknown times "height $= - 60$ " satisfy karte hain: ek physical (positive), ek unphysical (negative, launch se pehle). Hum positive root rakhte hain. $t = \frac{3 + 9 + 48}{2} = \frac{3 + 57}{2} = \frac{3 + 7.550}{2} = 5.275 s$ $x = u_{x} t = 20 \times 5.275 = 105.5 m$ Figure s03 upar yeh cliff arc overlay karta hai; horizontal reach level-ground formula se kaafi zyada hai.

L4·Q2 — Wall clear karna

Ek ball $u = 20 m/s$ , $θ = 4 5^{\circ}$ par kick ki gayi. $30 m$ door $3 m$ uchi wall hai. Kya ball usse clear karti hai? ( $sin 4 5^{\circ} = cos 4 5^{\circ} = 0.707$ use karo.)

Recall Solution

WHY trajectory equation? Hum given horizontal position $x = 30$ par height $y$ jaanna chahte hain, given time par nahi. Trajectory equation directly $x$ ke terms mein $y$ deta hai, isliye humein kabhi $t$ ki zaroorat nahi. Yeh directly parametric pair se aata hai: $x = u cos θ t$ se $t = \frac{x}{u cos θ}$ solve karo, phir $y = u sin θ t - \frac{1}{2} g t^{2}$ mein substitute karo. Pehla term $u sin θ \cdot \frac{x}{u cos θ} = x tan θ$ ban jaata hai, aur doosra $\frac{1}{2} g \cdot \frac{x ^{2}}{u ^{2} cos ^{2} θ}$ ban jaata hai, jo deta hai: $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}$ Ab $tan 4 5^{\circ} = 1$ , $cos^{2} 4 5^{\circ} = 0.5$ , $u^{2} = 400$ plug in karo: $y = 30 (1) - \frac{10 \times 3 0 ^{2}}{2 \times 400 \times 0.5} = 30 - \frac{9000}{400} = 30 - 22.5 = 7.5 m$ Ball wall tak pahunchne par $7.5 m$ par hai, jo $3 m$ se upar hai ⇒ wall clear ho jaati hai $4.5 m$ se.

Level 5 — Mastery

Ek relationship prove karo ya optimise karo.

L5·Q1 — Range–height link prove karo

Dikhao ki level ground ke liye, $R = \frac{4 H}{tan θ}$ .

Recall Solution

WHAT: $R$ aur $H$ dono ko same symbols se express karo aur divide karo. $R = \frac{u ^{2} s i n 2 θ}{g} = \frac{2 u ^{2} s i n θ c o s θ}{g}, H = \frac{u ^{2} s i n ^{2} θ}{2 g}$ Ratio banao ( $u^{2} / g$ cancel ho jaata hai): $\frac{R}{H} = \frac{2 s i n θ c o s θ}{\frac{1}{2} s i n ^{2} θ} = \frac{4 c o s θ}{s i n θ} = \frac{4}{t a n θ}$ Isliye $R = \frac{4 H}{tan θ}$ . Sanity check $θ = 4 5^{\circ}$ par: $tan 4 5^{\circ} = 1 \Rightarrow R = 4 H$ , matlab max-range angle par arc apni height se chaar guna wide hai.

L5·Q2 — Calculus se $T$ aur $H$ derive karo (koi memorised formula nahi)

Sirf $a_{y} = - g$ aur launch data $u = 30 m/s$ , $θ = 3 0^{\circ}$ se start karke, integration use karke $v_{y} (t)$ , $y (t)$ obtain karo, phir $T$ aur $H$ scratch se nikalo.

Recall Solution

Step 1 — acceleration ko velocity mein integrate karo. WHY integrate? $a_{y} = d v_{y} / d t$ , toh derivative undo karne se velocity milti hai: $v_{y} (t) = \int (- g) d t = - g t + C .$ $t = 0$ par, $v_{y} = u sin θ = 30 \times 0.5 = 15$ , toh $C = 15$ . Isliye $v_{y} = 15 - 10 t$ . Step 2 — velocity ko position mein integrate karo. $y (t) = \int (15 - 10 t) d t = 15 t - 5 t^{2} (origin at launch, C = 0) .$ Step 3 — time of flight: land hona jab $y = 0$ : $0 = t (15 - 5 t) \Rightarrow t = 0 or t = 3 s \Rightarrow T = 3 s .$ Step 4 — max height: top jab $v_{y} = 0$ : $15 - 10 t = 0 \Rightarrow t = 1.5 s$ . Plug in: $H = 15 (1.5) - 5 (1.5)^{2} = 22.5 - 11.25 = 11.25 m .$ Yeh $T = \frac{2 u s i n θ}{g} = \frac{2 \times 15}{10} = 3$ aur $H = \frac{u _{y}^{2}}{2 g} = \frac{225}{20} = 11.25$ se match karte hain. ✓

L5·Q3 — Cliff-edge shot ke liye optimal angle (challenge)

Ground level se tum $u = 20 m/s$ par maximum range chahte ho. Confirm karo ki $θ = 4 5^{\circ}$ optimal hai, phir maximum range batao.

Recall Solution

WHY $4 5^{\circ}$ : $R = \frac{u ^{2} sin 2 θ}{g}$ tab largest hai jab $sin 2 θ$ largest ho. Sine function $1$ par max hota hai jab uska argument $9 0^{\circ}$ ho, toh $2 θ = 9 0^{\circ} \Rightarrow θ = 4 5^{\circ}$ . $R_{m a x} = \frac{u ^{2}}{g} = \frac{2 0 ^{2}}{10} = 40 m .$ Koi bhi deviation (jaise $3 0^{\circ}$ : $sin 6 0^{\circ} = 0.866$ , $R = 34.6 m$ ) kam padta hai — peak confirm karta hai.

Recall One-line answer key (sirf numbers)

L1: $u_{x} = 40, u_{y} = 30$ ; $T = 6 s$ ; $H = 45 m, R = 240 m$ . L2: $v = 12.39 m/s$ at $36. 2^{\circ}$ ; $t = 4 s, x = 60 m$ ; $R = 57.85 m$ . L3: $θ = 3 0^{\circ}, R = 138.6 m$ ; dono $3 s$ mein land, gap $36 m$ ; $y = 11.25 m$ . L4: $t = 5.275 s, x = 105.5 m$ ; wall par height $= 7.5 m$ (clear hoti hai). L5: $R = 4 H / tan θ$ ; $T = 3 s, H = 11.25 m$ ; $R_{m a x} = 40 m$ .