JEE Main Projectile Motion: (45°+α) vs (45°−α) Concept 💡

❓ Concept

🎬 Projectile at (45° ± α) — Time of Flight Ratio in 59 Sec

Dono projectiles same speed se chhode gaye…
bas angle ek thoda upar, ek thoda neeche —
phir bhi time of flight alag kyun hota hai? ⏳🤔

👉 Reason simple hai: Time of flight sinθ par depend karta hai,
na ki range wale $\sin 2\theta$ par.

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🖼️ Concept Image

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✍️ Short Explanation

Is concept ko pakad liya,
toh JEE ke symmetry-based projectile questions seconds mein clear ho jaate hain 😎
Bas yaad rakho: time = vertical motion ka game.

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🔹 Step 1 — Time of Flight Formula (FOUNDATION 💯)**

Ground-to-ground projectile ke liye:

$$T=\frac{2u\sin \theta }{g}$$

📌 Key takeaway:

• $u$ same hai

• $g$ same hai

• Time of flight sirf $\sin\theta$ par depend karta hai

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🔹 Step 2 — Angles Given

$${\theta }_1={45}^{\circ }+\alpha ,{\theta }_2={45}^{\circ }-\alpha$$

So:

$$T_1\propto \sin ({45}^{\circ }+\alpha ),T_2\propto \sin ({45}^{\circ }-\alpha )$$

👉 Ratio nikaalne ke liye constants cancel ho jaate hain.

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🔹 Step 3 — Golden Trig Identity (VERY EXAM-FRIENDLY 🔥)**

Is pair ko yaad rakho:

$$\sin(45^\circ+\alpha)=\frac{\sin\alpha+\cos\alpha}{\sqrt{2}}$$ $$\sin ({45}^{\circ }-\alpha )=\frac{\cos \alpha -\sin \alpha }{\sqrt{\mathrm{2}}}$$

📌 Ratio mein $\sqrt{2}$ cancel ho jaata hai — kaam aur easy 👍

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🔹 Step 4 — Shortcut Expression (RESULT FORM ⭐)**

Time of flight ka ratio:

$$\frac{T_1}{T_2} = \frac{\sin(45^\circ+\alpha)}{\sin(45^\circ-\alpha)} = \frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}$$

👉 JEE options aksar:

• $\tan\alpha$ form

• ya $\sin 2\alpha$ related form
  mein hote hain — yahin se match ho jaata hai 😄

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🔹 Step 5 — Symmetry Insight (CHECK YOUR ANSWER 🧠)**

• Agar $\alpha=0$:

  • Dono angles same

  • $\Rightarrow T_1=T_2$

  • $\Rightarrow$ Ratio = 1 ✔️

• Jaise-jaise $\alpha$ badhta hai:

  • $45^\circ+\alpha$ wala projectile zyada vertical

  • Vertical component zyada

  • Time of flight zyada ⏳

📌 Intuition aur math perfectly match karte hain — good sign!

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⭐ Golden Summary Box

• Time of flight $\propto \sin\theta$
  

• Angles $45^\circ\pm\alpha$ ⇒ sine pair

• Ratio = ratio of sines

• Use $\sin(45^\circ\pm\alpha)$ identities

• Higher angle ⇒ longer time

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✅ Final Takeaway

🧠 One-line memory trick

Range = $\sin 2\theta$, Time = $\sin\theta$

Is difference ko samajh liya,
👉 Projectile ke symmetry questions JEE mein free marks ban jaate hain 🚀

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