1.1.20
Ek projectile bas do independent 1D motions ko glue karke bana hota hai :
Horizontal: constant velocity (koi force nahi, air ignore kar rahe hain). Boring, steady.Vertical: g
Inhe connect karne wali sirf ek cheez hai — time t har projectile formula generate karta hai. Hum kabhi memorize nahi karte — derive karte hain.
Components mein kyun split karte hain? Kyunki force (gravity) sirf vertically act karta hai. Horizontal motion par koi force nahi, isliye woh uniform honi chahiye. Split karne se hum har axis par simple 1D equations use kar sakte hain.
Kya hai: Hawa mein total time, launch se lekar same height par wapas land karne tak.
Kaise: Vertical displacement y = u y t − 2 1 g t 2 y = 0
0 = u sin θ t − 2 1 g t 2 = t ( u sin θ − 2 1 g t )
Yeh step kyun? t t = 0
Intuition Symmetry insightTop tak pahunchne ka time = g u sin θ = = 2 ×
Kya hai: Sabse ucha vertical point, jahan v y = 0
Kaise: v y 2 = u y 2 − 2 g H v y = 0
0 = ( u sin θ ) 2 − 2 g H
Yeh step kyun? v y 2 = u y 2 − 2 g y time-free energy-jaisi relation hai — perfect hai kyunki peak par hum velocity jaante hain (0
Worked example Time use karke cross-checkt = g u s i n θ y = u y t − 2 1 g t 2 H = u sin θ ⋅ g u s i n θ − 2 1 g g 2 u 2 s i n 2 θ = g u 2 s i n 2 θ − 2 g u 2 s i n 2 θ = 2 g u 2 s i n 2 θ ✓ Yeh step kyun? Do independent routes se same answer milna derivation ko prove karta hai.
Kya hai: Poori flight T
Kaise: Horizontal motion uniform hai, isliye R = u x ⋅ T
R = u cos θ ⋅ g 2 u s i n θ = g 2 u 2 s i n θ c o s θ
Identity 2 sin θ cos θ = sin 2 θ
Yeh step kyun? Range = (constant horizontal speed) × (hawa mein total time). Hawa mein time poori tarah vertical motion se set hota hai — yahi coupling hai.
4 5 ∘ R ∝ sin 2 θ 2 θ = 9 0 ∘ ⇒ θ = 4 5 ∘ Maximum range = g u 2 θ 9 0 ∘ − θ same range dete hain kyunki sin 2 θ = sin ( 18 0 ∘ − 2 θ )
T = g 2 u s i n θ , H = 2 g u 2 s i n 2 θ , R = g u 2 s i n 2 θ H sin 2 R sin 2 θ T sin Sab θ = 0 θ = 0 T = 0 , H = 0 , R = 0
Worked example Example 1 — full kitEk ball u = 20 m/s θ = 3 0 ∘ g = 10 m/s 2
Time of flight: T = 10 2 ( 20 ) ( 0.5 ) = 2 s Kyun? sin 3 0 ∘ = 0.5
Max height: H = 2 ( 10 ) ( 20 ) 2 ( 0.5 ) 2 = 20 400 ⋅ 0.25 = 5 m Kyun? Sine aur speed dono ko square karo.
Range: R = 10 ( 20 ) 2 sin 6 0 ∘ = 10 400 ⋅ 0.866 ≈ 34.6 m Kyun? sin 2 θ = sin 6 0 ∘ sin 3 0 ∘
Worked example Example 2 — Forecast-then-VerifySame speed, lekin θ = 6 0 ∘ Forecast: Range Example 1 ke identical honi chahiye (complementary angles).
Verify: R = 10 400 sin 12 0 ∘ = 10 400 ( 0.866 ) ≈ 34.6 m H = 20 400 ( 0.75 ) = 15 m T = 10 2 ( 20 ) ( 0.866 ) ≈ 3.46 s
Common mistake Range meinsin 2 θ sin θ Kyun sahi lagta hai: Baaki har formula sin θ Fix: Range aata hai u cos θ × g 2 u s i n θ sin cos collapse hokar sin 2 θ 2 sin θ cos θ
Common mistake Time of flight mein factor of 2 bhool jaanaKyun sahi lagta hai: t up = g u s i n θ Fix: Yeh sirf rise time hai. Total flight = upar aur neeche = 2 t up t up
Common mistake Horizontal velocity ko changing treat karnaKyun sahi lagta hai: "Sab kuch eventually slow down ho jaata hai."
Fix: Koi horizontal force nahi ⇒ u x u y
Common mistake Range formula tab use karna jab launch aur landing heights alag honKyun sahi lagta hai: Yahi formula tumne memorize kiya hai.
Fix: R = g u 2 s i n 2 θ same height assume karta hai. Ek cliff se, T y = 0
Recall Ek 12-saal ke bacche ko explain karo (click to reveal)Socho tum ek ball throw kar rahe ho. Do cheezein ek saath lekin alag alag hoti hain: woh aage steady pace se travel karti hai (kuch bhi use sideways nahi dhakelta), aur woh *upar jaati hai phir neeche kyunki gravity use ek invisible rubber band ki tarah wapas kheenchti hai. Hawa mein woh kitna time spend karti hai woh sirf upar-neeche wale part par depend karta hai. Jab tak woh wahan float kar rahi hai, aage wala part steadily move karta rehta hai. Toh:
Hawa mein time = upar-neeche ki trip kitni der leti hai.Kitna ucha = upar ki taraf throw kitni strong thi.Kitna door = forward speed × hawa mein time.
Exactly 45° ke slant par throw karna use sabse door bhejta hai, kyunki tum "kaafi der tak upar rehna" aur "tezi se aage jaana" dono ko balance karte ho.
u g "Time fair hai, Height steep hai, Range loose hai."
T ime: u 1 , g 1 H eight: u 2 , g 1 2 se divide karo (steep climb ki cost hoti hai).R ange: u 2 , g 1 sin 2 θ
Projectile motion mein kaun se do motions hote hain, aur unhe kya link karta hai? Horizontal (uniform velocity) aur vertical (free fall under g t
y = u sin θ t − 2 1 g t 2 = 0 t ( u sin θ − 2 1 g t ) = 0 T = g 2 u s i n θ
Time of flight ka formula? T = g 2 u s i n θ
Maximum height ka formula? H = 2 g u 2 s i n 2 θ
Horizontal range ka formula? R = g u 2 s i n 2 θ
Max range 45° par kyun hoti hai? R ∝ sin 2 θ 2 θ = 90° θ = 45° = u 2 / g
Complementary angles same range kyun dete hain? sin 2 θ = sin ( 180° − 2 θ ) θ 90° − θ R
Max height par vertical velocity kya hoti hai? Zero (momentarily rise karna band ho jaata hai); horizontal velocity abhi bhi u cos θ
2 sin θ cos θ 2 sin θ cos θ = sin 2 θ
u = 20 , θ = 30° , g = 10 T = 2 H = 5 R ≈ 34.6
Horizontal ux = u cos theta
Vertical uy = u sin theta
Time of flight T = 2u sin theta / g
Max height H = u squared sin sq theta / 2g
Range R = u squared sin 2theta / g
Identity 2 sin cos = sin 2theta
