Projectile motion — horizontal - vertical independence, full derivation: Worked examples

1.1.19 · D3 · Physics › Measurement, Vectors & Kinematics › Projectile motion — horizontal - vertical independence, full

Yeh page ek drill hai. Parent note ne machinery banai; yahan hum har tarah ki ball uss par throw karte hain. Plan yeh hai: pehle saare case classes map karo jo ek projectile problem mein ho sakti hain, phir har case ka ek example solve karo taaki koi bhi scenario unfamiliar na lage.

Scenario matrix

Har projectile problem inhi cells mein se ek hai. Har row ek tarika hai jisme setup alag ho sakta hai; neeche jo examples follow karte hain unhe uss cell ke saath tag kiya gaya hai jo woh cover karte hain.

Cell	Kya khaas baat hai	Danger jo chupi hai	Example
A Level launch, generic $θ$	launch height = landing height	koi nahi — $T, H, R$ mein plug karo	Ex 1
B $θ = 0$ (horizontal throw)	shuruaat se hi $u_{y} = 0$	$R = u^{2} sin 2 θ / g$ use nahi kar sakte	Ex 2
C $θ = 9 0^{\circ}$ (seedha upar)	$u_{x} = 0$ , range zero hai	yeh actually 1-D hai	Ex 3
D Landing launch se neeche (cliff)	$y_{final} < 0$	naive $T, R$ formulas fail ho jaate hain	Ex 4
E Landing launch se upar (wall/roof par)	$y_{final} > 0$	do candidate times	Ex 5
F Complementary angles	$θ$ aur $9 0^{\circ} - θ$	same range, alag $T, H$	Ex 6
G Flight ke dauran velocity/direction	time $t$ par $v_{x}, v_{y}$ chahiye	$v_{y}$ ka sign top par flip hota hai	Ex 7
H Real-world word problem	hidden numbers, unit traps	reading, physics nahi	Ex 8
I Exam twist — $θ$ solve karo	$R$ given, angle nikalo	$sin$ do answers deta hai	Ex 9
J Degenerate / limit check	$u = 0$ , $g \to$ large, $θ \to 0$	formulas sane rehne chahiye	Ex 10

Cell A — Level launch, generic angle

Worked example Example 1 — vanilla case

Ek ball ground se $u = 20 m/s$ , $θ = 3 0^{\circ}$ , $g = 10 m/s^{2}$ par nikali jaati hai. Flight time $T$ , max height $H$ , range $R$ , aur landing par speed nikalo.

Forecast: compute karne se pehle andaaza lagao — kya landing speed $20 m/s$ se zyada, kam, ya barabar hogi?

Launch velocity ko split karo. $u_{x} = 20 cos 3 0^{\circ} = 20 (0.866) = 17.32 m/s$ , $u_{y} = 20 sin 3 0^{\circ} = 20 (0.5) = 10 m/s$ . Yeh step kyun? Dono axes independent hain; hume har ek ko uski apni starting speed deni padti hai.
Flight time $T = \frac{2 u _{y}}{g} = \frac{2 ( 10 )}{10} = 2 s$ . Yeh step kyun? Ball $y = 0$ par wapas aati hai jab vertical motion apna up-and-down poora kare; sirf $u_{y}$ yeh decide karta hai.
Max height $H = \frac{u _{y}^{2}}{2 g} = \frac{100}{20} = 5 m$ . Yeh step kyun? Top par $v_{y} = 0$ ; sirf vertical piece ki energy/kinematics $H$ deti hai.
Range $R = u_{x} T = 17.32 (2) = 34.64 m$ . Yeh step kyun? Horizontal speed constant hai, isliye distance = speed × total time.
Landing speed. Symmetry se $v_{x} = 17.32$ , $v_{y} = - 10$ (same magnitude, ab neeche). Speed $= 17.3 2^{2} + 1 0^{2} = 300 + 100 = 20 m/s$ .

Verify: landing speed = launch speed $20 m/s$ — sahi, kyunki level ground par gravity exactly utni hi kinetic energy wapas karti hai jo usne li thi. Forecast answer: barabar. Units: m/s ✓.

Cell B — Horizontal throw ( $θ = 0$ )

Worked example Example 2 — drop vs shoot

Ek patthar $20 m$ ki cliff se horizontally $u = 15 m/s$ par throw kiya jaata hai. Woh kab aur kahan land karta hai? Ek simply drop kiye patthar se compare karo.

Forecast: kya throw kiya hua patthar ground par pehle girega, kyunki uske saath "zyada kuch ho raha hai"?

Resolve karo. $θ = 0 \Rightarrow u_{x} = 15$ , $u_{y} = 0$ . Yeh step kyun? Flat throw apni saari speed horizontal mein daalta hai; vertical rest se shuru hoti hai.
Vertical fall. Yahan neeche ko positive lete hain: $20 = \frac{1}{2} (10) t^{2} \Rightarrow t^{2} = 4 \Rightarrow t = 2 s$ . Yeh step kyun? Vertically patthar ek pure free-fall object hai; horizontal speed girne ke liye irrelevant hai.
Horizontal distance. $x = u_{x} t = 15 (2) = 30 m$ . Yeh step kyun? Shared clock $t$ ke across constant horizontal velocity.
Dropped patthar. $20 = \frac{1}{2} (10) t^{2} \Rightarrow t = 2 s$ — bilkul same. Yeh step kyun? Drop kiye patthar ka bhi $u_{y} = 0$ hai aur woh same $20 m$ same $g$ ke neeche girta hai; uski vertical equation byte-for-byte throw kiye patthar jaisi hai, isliye same $t$ dena zaroori hai. Yeh independence ka visible proof hai.

Verify: dono $2 s$ lete hain; horizontal throw fall time nahi badlata — independence ka yahi poora point hai. Forecast answer: nahi, dono saath girtey hain. Units ✓.

Cell C — Seedha upar ( $θ = 9 0^{\circ}$ )

Worked example Example 3 — degenerate vertical shot

Ek ball seedhi upar $u = 30 m/s$ par fire ki jaati hai. $T$ , $H$ , aur $R$ nikalo.

Forecast: purely vertical shot ke liye range kya honi chahiye?

Resolve karo. $cos 9 0^{\circ} = 0 \Rightarrow u_{x} = 0$ ; $sin 9 0^{\circ} = 1 \Rightarrow u_{y} = 30$ . Yeh step kyun? Saari speed upar jaati hai; kuch sideways nahi.
Flight time $T = \frac{2 ( 30 )}{10} = 6 s$ . Yeh step kyun? Ball upar jaati hai, rukti hai, aur launch height par wapas aati hai; vertical motion ka up-and-down airborne time set karta hai, isliye $T = 2 u_{y} / g$ abhi bhi apply hota hai.
Max height $H = \frac{3 0 ^{2}}{2 ( 10 )} = \frac{900}{20} = 45 m$ .
Range $R = u_{x} T = 0 (6) = 0 m$ . Yeh step kyun? Koi horizontal speed nahi ⇒ woh launch point par wapas aati hai. Yeh actually disguise mein ek 1-D problem hai.

Verify: $R = 0$ — "projectile" formulas gracefully straight-up case mein collapse ho jaate hain. Forecast answer: zero. Units ✓.

Cell D — Launch se neeche landing (cliff)

Worked example Example 4 — cliff se angle par launch

$25 m$ cliff ke top se, ek ball $u = 20 m/s$ , $θ = 3 0^{\circ}$ horizontal ke upar par throw ki jaati hai. Woh kitni der hawaon mein hai, aur cliff base se kitni door land karti hai?

Forecast: kya flight time Example 1 ke level-ground $T = 2 s$ se zyada hoga ya kam?

Neeche di figure setup dikhati hai: ball cliff top se launch hoti hai, launch level se upar arc karti hai, phir base tak extra $25 m$ girती hai. Notice karo ki red arc dashed launch line ke neeche extend hoti hai — woh extra fall exactly wahi hai jo naive $T$ formula yahan galat karta hai.

Resolve karo. $u_{x} = 20 cos 3 0^{\circ} = 17.32$ , $u_{y} = 20 sin 3 0^{\circ} = 10$ .
Vertical equation ko true landing height par set karo. Cliff ka base launch se $25 m$ neeche hai, isliye $y = - 25$ : $- 25 = 10 t - \frac{1}{2} (10) t^{2} \Rightarrow 5 t^{2} - 10 t - 25 = 0 \Rightarrow t^{2} - 2 t - 5 = 0.$ Yeh step kyun? $T = 2 u sin θ / g$ formula assume karta hai landing launch height par hogi — yahan nahi hoti, isliye hum raw $y (t)$ equation par wapas jaate hain.
Quadratic solve karo. $t = \frac{2 \pm 4 + 20}{2} = 1 \pm 6$ . Positive root: $t = 1 + 2.449 = 3.449 s$ . Yeh step kyun? Sirf positive time physical hai (negative root parabola ka "past" hai).
Horizontal distance. $x = u_{x} t = 17.32 (3.449) = 59.73 m$ . Yeh step kyun? Ab jo vertical motion ne hume bataya hai kab land hoti hai, us instant par horizontal position sirf constant speed × us time hai — shared clock $t$ dono axes ko link karta hai.

Verify: ball zyada der hawaon mein rehti hai ( $3.45 s > 2 s$ ) kyunki woh launch level se aage girती rehti hai — figure ke extended red arc se match karta hai. Forecast answer: zyada. Units ✓.

Cell E — Launch se upar landing (wall par)

Worked example Example 5 — wall clear karna (do candidate times)

Ek ball ground se $u = 25 m/s$ , $θ = 5 3^{\circ}$ ( $sin 5 3^{\circ} = 0.8$ , $cos 5 3^{\circ} = 0.6$ ) par launch ki jaati hai. Aage kahin ek $15 m$ height ki wall hai. Kis-kis time par ball $15 m$ height par hai?

Forecast: ek projectile apni peak se neeche kisi given height par kitne distinct times par ho sakta hai?

Resolve karo. $u_{x} = 25 (0.6) = 15$ , $u_{y} = 25 (0.8) = 20$ .
$y = 15$ set karo. $15 = 20 t - 5 t^{2} \Rightarrow 5 t^{2} - 20 t + 15 = 0 \Rightarrow t^{2} - 4 t + 3 = 0$ . Yeh step kyun? Hume har woh instant chahiye jab ball us height par ho, isliye poora quadratic solve karo — sirf ek root nahi.
Factor karo. $(t - 1) (t - 3) = 0 \Rightarrow t = 1 s$ (upar jaate hue) aur $t = 3 s$ (neeche aate hue). Yeh step kyun? Ek projectile har sub-peak height par do baar guzarta hai — ek baar rise mein, ek baar fall mein. Dono roots real aur physical hain.
Peak check. Max height $H = \frac{u _{y}^{2}}{2 g} = \frac{2 0 ^{2}}{2 ( 10 )} = 20 m > 15 m$ , isliye $15 m$ actually reachable hai. Yeh step kyun? Agar wall $H$ se oocha hoti toh quadratic ke koi real roots nahi hote (ball wahan pahunchti hi nahi); $H$ compute karna confirm karta hai ki hamare do times genuine hain, algebraic ghosts nahi.

Verify: dono times $t_{top} = \frac{u _{y}}{g} = 2 s$ ke baare mein symmetric hain; wakai $\frac{1 + 3}{2} = 2$ ✓. Forecast answer: do (agar peak se neeche ho). Units ✓.

Cell F — Complementary angles

Worked example Example 6 — same range, baaki sab alag

Ek cannon $u = 40 m/s$ par fire karti hai, ek baar $θ = 3 0^{\circ}$ par aur ek baar $θ = 6 0^{\circ}$ par. $R$ , $T$ , aur $H$ compare karo.

Forecast: dono shots ke liye teeno quantities mein se kaun si barabar hain?

Figure dono arcs overlay karta hai. Notice karo ki woh ground par same landing point par milte hain (equal range, yellow line) even though $6 0^{\circ}$ arc $3 0^{\circ}$ wale se kahin zyada upar jaata hai — "same $R$ , different $H$ " ki ek striking picture.

Ranges. $R = \frac{u ^{2} sin 2 θ}{g}$ . $3 0^{\circ}$ ke liye: $sin 6 0^{\circ} = 0.866$ , $R = \frac{1600 ( 0.866 )}{10} = 138.6 m$ . $6 0^{\circ}$ ke liye: $sin 12 0^{\circ} = 0.866$ , same $R = 138.6 m$ . Yeh step kyun? $sin 2 θ = sin (18 0^{\circ} - 2 θ)$ , isliye complementary angles $sin 2 θ$ share karte hain — equal range.
Times. $T = \frac{2 u sin θ}{g}$ . $3 0^{\circ}$ : $T = \frac{2 ( 40 ) ( 0.5 )}{10} = 4 s$ . $6 0^{\circ}$ : $T = \frac{2 ( 40 ) ( 0.866 )}{10} = 6.93 s$ . Yeh step kyun? $T \propto sin θ$ , aur $sin 6 0^{\circ} > sin 3 0^{\circ}$ — steeper shot zyada der rukti hai.
Heights. $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ . $3 0^{\circ}$ : $\frac{1600 ( 0.25 )}{20} = 20 m$ . $6 0^{\circ}$ : $\frac{1600 ( 0.75 )}{20} = 60 m$ . Yeh step kyun? $H = \frac{u _{y}^{2}}{2 g}$ with $u_{y} = u sin θ$ deta hai $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ ; kyunki $H$ $sin^{2} θ$ par depend karta hai, steeper $6 0^{\circ}$ shot (bada $sin θ$ ) kaafi zyada upar chadhti hai.

Verify: ranges barabar (138.6 m each) ✓; $6 0^{\circ}$ shot 3× zyada upar jaati hai aur zyada der rehti hai — tall vs flat arcs dekho. Forecast answer: sirf range barabar hai. Units ✓.

Cell G — Flight ke dauran velocity aur direction

Worked example Example 7 — mid-flight mein speed aur direction

Example 1 ki ball ke liye ( $u_{x} = 17.32$ , $u_{y} = 10$ , $g = 10$ ), $t = 1.5 s$ par velocity vector nikalo: uske components, speed, aur horizontal ke saath angle.

Forecast: $t = 1.5 s$ par ( $t = 1 s$ peak ke baad) kya ball upar ja rahi hai ya neeche?

Horizontal velocity. $v_{x} = u_{x} = 17.32 m/s$ (constant, unchanged). Yeh step kyun? $a_{x} = 0$ ; horizontal speed kabhi nahi badalti — classic trap.
Vertical velocity. $v_{y} = u_{y} - g t = 10 - 10 (1.5) = - 5 m/s$ . Yeh step kyun? Negative matlab neeche — ball apni peak ke past hai (peak $t = 1 s$ par thi).
Speed. $∣ v ∣ = v_{x}^{2} + v_{y}^{2} = 17.3 2^{2} + 5^{2} = 300 + 25 = 325 = 18.03 m/s$ . Yeh step kyun? Velocity ek vector hai; uske do perpendicular components ek right triangle ki legs banate hain, isliye actual speed hypotenuse hai (Pythagoras) — $v_{x}$ aur $v_{y}$ simply add nahi kar sakte.
Horizontal ke neeche direction. $α = arctan \frac{∣ v _{y} ∣}{v _{x}} = arctan \frac{5}{17.32} = 16. 1^{\circ}$ horizontal ke neeche. Yeh step kyun? Velocity vector ka angle uske components ke right triangle par opposite/adjacent hai; $arctan$ poochta hai "is tangent wala angle kaun sa hai?" Dekho Vectors — Resolving into Components.

Verify: $v_{y} < 0$ descent confirm karta hai; speed $18.03 < 20 m/s$ (launch) kyunki woh abhi launch height par wapas nahi pahunchi. Forecast answer: neeche. Units ✓.

Cell H — Real-world word problem

Worked example Example 8 — water fountain

Ek garden fountain paani $u = 14 m/s$ , $θ = 4 5^{\circ}$ par shoot karta hai. Gardener ek circular pool chahta hai jo saara paani pakad le. Minimum pool radius (nozzle se measure) kitni chahiye? ( $g = 9.8 m/s^{2}$ .)

Forecast: kya sabse bada splash $4 5^{\circ}$ par land hota hai ya koi aur angle zyada door tak pahunchega?

Hidden ask identify karo. "Saara paani pakad le" = jet ki range. Yeh step kyun? Word problems physics quantity ko plain language mein chupaate hain; range wahi hai jo "kitni door land hota hai" matlab hai.
Range formula use karo. $R = \frac{u ^{2} sin 2 θ}{g} = \frac{1 4 ^{2} sin 9 0 ^{\circ}}{9.8} = \frac{196 ( 1 )}{9.8} = 20 m$ . Yeh step kyun? Level ground (nozzle aur pool same height par) ⇒ standard $R$ apply hota hai, aur $θ = 4 5^{\circ}$ $sin 2 θ = 1$ banata hai.
Answer. Radius $= 20 m$ .

Verify: $4 5^{\circ}$ max-range angle hai, isliye yeh worst case radius bhi hai — chhote angles paas mein land karte hain. Units: $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ ✓. Forecast answer: $4 5^{\circ}$ sabse door hai.

Cell I — Exam twist: angle ke liye solve karo

Worked example Example 9 — range given, launch angle nikalo

$u = 30 m/s$ par throw ki gayi ball level ground par $45 m$ door land karti hai ( $g = 10$ ). Kaun se launch angle(s) yeh achieve karte hain?

Forecast: kitne launch angles same jagah hit kar sakte hain?

$θ$ ke liye range equation set karo. $45 = \frac{3 0 ^{2} sin 2 θ}{10} = 90 sin 2 θ \Rightarrow sin 2 θ = 0.5$ . Yeh step kyun? Ab hume $R$ aur $u$ pata hai; unknown $sin 2 θ$ ke andar hai, isliye use isolate karo.
Sine invert karo. $2 θ = arcsin (0.5) = 3 0^{\circ}$ ya $2 θ = 18 0^{\circ} - 3 0^{\circ} = 15 0^{\circ}$ . Yeh step kyun? $sin$ ek angle aur uske supplement par same value leta hai — yahi do solutions ka source hai.
Solve karo. $θ = 1 5^{\circ}$ ya $θ = 7 5^{\circ}$ . Yeh step kyun? Yeh complementary hain ( $1 5^{\circ} + 7 5^{\circ} = 9 0^{\circ}$ ) — exactly Cell F pattern.

Verify: $θ = 1 5^{\circ}$ plug karo: $sin 3 0^{\circ} = 0.5$ , $R = 90 (0.5) = 45 m$ ✓; $7 5^{\circ}$ ke liye bhi same. Forecast answer: do (ek low aur ek high shot). Units ✓.

Cell J — Degenerate & limiting checks

Worked example Example 10 — sanity limits (kya maths honest rehti hai?)

Teen extremes par formulas test karo aur confirm karo ki woh sahi behave karte hain.

Forecast: agar hum $g$ badhayein toh kya $R$ badhega ya ghutega?

Zero speed, $u = 0$ . $R = \frac{0 \cdot sin 2 θ}{g} = 0$ , $H = \frac{0}{2 g} = 0$ , $T = \frac{0}{g} = 0$ . Yeh step kyun? Koi launch nahi ⇒ koi motion nahi; ek well-behaved formula ko har quantity ke liye exactly zero return karna chahiye, aur har ek karta hai.
Angle $\to 0$ . Jaise $θ \to 0$ , $sin θ \to 0 \Rightarrow T = \frac{2 u sin θ}{g} \to 0$ aur $H = \frac{u ^{2} sin ^{2} θ}{2 g} \to 0$ ; aur $R = \frac{u ^{2} sin 2 θ}{g} \to 0$ bhi. Truly flat ground par ground level se horizontal throw ke paas kahan rise karne ki jagah nahi, isliye woh "instantly land karta hai." Yeh step kyun? Yeh Cell A (angled level launch) aur Cell B (horizontal throw) ke beech ki boundary check karta hai: flat ground par horizontal case degenerate ho jaata hai, yahi exactly wajah hai Cell B ko interesting hone ke liye cliff chahiye.
Stronger gravity, $g \to$ large. $u = 20$ , $θ = 4 5^{\circ}$ ke saath: $g = 10 \Rightarrow R = \frac{400 ( 1 )}{10} = 40 m$ ; $g = 20 \Rightarrow R = \frac{400}{20} = 20 m$ . Yeh step kyun? $g$ double karne se range half ho jaati hai — us intuition se match karta hai ki zyada pull karne wala planet ball ko jaldi neeche kheench leta hai, flight short kar deta hai.

Verify: saare degenerate outputs finite aur non-negative hain; $g$ badhne par $R$ ghutaa ( $40 \to 20 m$ ), aur har $u = 0$ ya $θ \to 0$ output $0$ mein collapse ho jaata hai jaise physics require karti hai. Forecast answer: ghutega. Units ✓.

Active recall

Recall Main kis cell mein hoon?

Ball rooftop se flat throw ki gayi → ::: Cell D (launch se neeche landing — poora $y (t)$ use karo, $T = 2 u sin θ / g$ nahi).
"Kis do times par woh height $h$ par hai?" → ::: Cell E (quadratic, do roots peak time ke baare mein symmetric).
$R$ aur $u$ given, $θ$ nikalo → ::: Cell I ( $sin 2 θ$ do complementary angles deta hai).
Ball seedhi upar fire ki gayi → ::: Cell C (range zero, actually 1-D).