Projectile Range, Height, and Angle

Create a high-quality 2D educational physics animation for JEE Main and JEE Advanced students explaining the relationship between angle of projection, horizontal range and maximum height of a projectile. Use a clean classroom/physics-animation style, white or dark neutral background, accurate mathematical diagrams, smooth vector animation, clearly labelled axes, projectile trajectories, arrows and equations. No photorealistic people. The focus must remain on physics concepts and visual understanding. Scene 1 — Satellite/Projectile Launch Introduction Show a spacecraft/satellite being launched from the surface of the Earth with an initial velocity u. Gradually transform the satellite-launch illustration into the standard projectile-motion model: an object projected from ground with velocity u at an angle \theta with the horizontal. Display: > u_x=u\cos\theta > > u_y=u\sin\theta > Explain visually that the horizontal and vertical components determine the trajectory. Scene 2 — Projectile Trajectory Show the projectile following a parabolic trajectory and landing at the same horizontal level from which it was launched. Display: > R=\frac{u^2\sin2\theta}{g} > > H_{\max}=\frac{u^2\sin^2\theta}{2g} > > T=\frac{2u\sin\theta}{g} > Keep u and g constant throughout the demonstration. Scene 3 — Range changes with angle Animate several projectiles launched with the same initial velocity but different angles: 15°, 30°, 45°, 60°, 75° Show each trajectory sequentially using the same starting point and landing level. Clearly demonstrate: From 0° to 45° → horizontal range increases. At 45° → horizontal range is maximum. From 45° to 90° → horizontal range decreases. Highlight the 45° trajectory as the maximum-range trajectory. Display: > R_{\max}=\frac{u^2}{g} > \qquad\text{at}\qquad > \theta=45^\circ > Scene 4 — Why 45° gives maximum range Zoom into the equation: > R=\frac{u^2\sin2\theta}{g} > Since u and g are constant, show that the range depends on \sin2\theta. Animate: > \sin2\theta\leq1 > Therefore maximum range occurs when: > 2\theta=90^\circ > > \boxed{\theta=45^\circ} > Make this visually prominent. Scene 5 — Complementary Angles Show two projectiles launched with the same initial speed at: > 30^\circ \quad\text{and}\quad60^\circ > Display: > \sin(2\times30^\circ)=\sin60^\circ > > \sin(2\times60^\circ)=\sin120^\circ > Since: > \sin60^\circ=\sin120^\circ > show both projectiles landing at the same horizontal position. Then demonstrate another pair: 20° and 70° and show that their ranges are again equal. Clearly label: Complementary angles → Same Range > \boxed{\theta_1+\theta_2=90^\circ} > Scene 6 — But the heights are different Compare the 30° and 60° trajectories. Show that although their horizontal ranges are identical, their maximum heights are different. Display: > H_{\max}=\frac{u^2\sin^2\theta}{2g} > Visually show the 60° projectile reaching a substantially greater height than the 30° projectile. Explain through animation: > \sin60^\circ>\sin30^\circ > therefore: > H_{60^\circ}>H_{30^\circ} > Scene 7 — Final Concept Summary End with a clean comparison table/infographic: ANGLE | RANGE | MAXIMUM HEIGHT 30° → Same as 60° → Lower 45° → Maximum → Intermediate 60° → Same as 30° → Higher Then display three key JEE concepts: 1. Range increases from 0° to 45°. 2. Maximum range occurs at 45°. 3. Complementary angles give equal range but different maximum heights. Finish with: > \boxed{R=\frac{u^2\sin2\theta}{g}} > > \boxed{R(\theta)=R(90^\circ-\theta)} > > \boxed{H_{\max}=\frac{u^2\sin^2\theta}{2g}} > Add a final “JEE Main & Advanced Concept” title card.