Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

❓ 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 θ \sin 2\theta  par. 🖼️ Concept Image ✍️ 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 . 🔹 Step 1 — Time of Flight Formula (FOUNDATION 💯)** Ground-to-ground projectile ke liye: T = 2 u sin ⁡ θ g 📌 Key takeaway: u u  same hai g g  same hai Time of flight sirf sin ⁡ θ \sin\theta  par depend karta hai 🔹 Step 2 — Angles Given θ 1 = 45 ∘ + α , θ 2 = 45 ∘ − α So: T 1 ∝ sin ⁡ ( 45 ∘ + α ) , T 2 ∝ sin ⁡ ( 45 ∘ − α ) 👉 Ratio nikaalne ke liye constants cancel ho jaate hain. 🔹 Step 3 — Golden Trig Identity (VERY...

  ❓ Question Two projectiles are fired from the ground with the same initial speed from the same point , at angles ( 45 ∘ + α ) and ( 45 ∘ − α ) with the horizontal. Find the ratio of their times of flight . 🖼️ Question Image ✍️ Short Solution This is a classic JEE symmetry question from projectile motion. The trick is to remember what depends on sine and what depends on sine double-angle . 🔹 Step 1 — Time of flight formula For a projectile fired from ground with speed u u u at angle θ \theta : T = 2 u sin ⁡ θ g 📌 Time of flight depends on sin ⁡ θ \sin\theta sin θ (not on sin ⁡ 2 θ \sin 2\theta sin 2 θ ). 🔹 Step 2 — Write times for both projectiles Let: T 1 T_1 ​ = time of flight at angle ( 45 ∘ + α ) (45^\circ+\alpha) T 2 T_2 ​ = time of flight at angle ( 45 ∘ − α ) Using the formula: T 1 = 2 u sin ⁡ ( 45 ∘ + α ) g T_1=\frac{2u\sin(45^\circ+\alpha)}{g} T 2 = 2 u sin ⁡ ( 45 ∘ − α ) g T_2=\frac{2u\sin(45^\circ-\alpha)}{g} 🔹 Step 3 — Take the r...