Projectile Motion Trick: Time of Flight Ratio in 60 Seconds! 🔥
❓ 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...