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Projectile Tracked by a Radar Station

A projectile launches from point O at 150 m/s and 75° above the horizontal, while a radar station sits 20 m away at point A. The animation splits the launch velocity into horizontal and vertical components, tracks the parabolic path to maximum height, and computes the time, vertical rise, and horizontal range. It then forms a right triangle between the radar and the projectile's peak to find the radar's line-of-sight distance and angle. Useful for physics students studying projectile motion and vector geometry.

16:9 · every frame verified for overlaps, spacing and edges before rendering

The prompt that made it

Create an educational 2D physics animation explaining a projectile motion problem step by step. Scene 1: Show a boat on a horizontal surface. Point O is the projectile launch point and point A is a radar located 20 meters horizontally from O. A projectile is launched from O with initial velocity v0 = 150 m/s at an angle of 75° above the horizontal. Show the x-axis horizontally and y-axis vertically. Label all points, distances, angles, and vectors clearly. Scene 2: Resolve the initial velocity into horizontal and vertical components: Vx0 = 150 cos(75°) = 38.8 m/s Vy0 = 150 sin(75°) = 144 m/s Animate the velocity vector splitting into its x and y components. Scene 3: Show the projectile moving upward along a parabolic trajectory until it reaches maximum height. At maximum height, show that the vertical velocity becomes zero: Vfy = 0 Calculate the maximum vertical displacement: Vfy² = Vy0² - 2gΔy 0 = 144² - 2(9.8)Δy Δy = 1057 m Then calculate the time to reach maximum height: Δy = Vy0 t - 1/2 gt² 1057 = 144t - 1/2(9.8)t² t = 15.13 s Scene 4: Calculate the horizontal displacement: Δx = Vx0 t Δx = 38.8(15.13) Δx = 587 m Because the radar point A is 20 m from O, the horizontal distance from A to the projectile is: 587 - 20 = 567 m Show a right triangle between radar A and the projectile at maximum height. The vertical side is 1057 m and the horizontal side is 567 m. Scene 5: Calculate the radial distance r: r = √[(1057)² + (567)²] r ≈ 1200 m Calculate the radar angle: θ = tan⁻¹(1057/567) θ ≈ 61° Animate the radar line from A to the projectile and label it r. Show the angle θ between the horizontal axis and the radial line. Scene 6: Explain the velocity components in polar coordinates (r-θ). At maximum height, the projectile velocity is horizontal: V = 38.8 m/s Radial velocity: Vr = V cos(61°) Vr = 18.8 m/s Tangential velocity: Vθ = V sin(61°) Vθ = 33.9 m/s Since: Vθ = r θ̇ 33.9 = 1057 θ̇ θ̇ = 0.028 rad/s Animate the velocity vector splitting into radial and tangential components. Scene 7: Show gravity acting vertically downward with acceleration g = 9.8 m/s². Resolve acceleration into radial and tangential components. The angle between the radial direction and downward gravity direction is 29°. ar = -9.8 cos(29°) ar = -8.57 m/s² aθ = 9.8 sin(29°) aθ = 4.75 m/s² Show the acceleration vector pointing downward and animate its components along the radial and tangential directions. Scene 8: Finally, calculate the radius of curvature. At maximum height, the velocity is horizontal and gravity is perpendicular to the velocity, so the normal acceleration is: an = 9.8 m/s² Use: an = V² / ρ 9.8 = (38.8)² / ρ ρ ≈ 153 m End the video with a clean summary screen showing: r = 1200 m θ = 61° Vr = 18.8 m/s Vθ = 33.9 m/s θ̇ = 0.028 rad/s ar = -8.57 m/s² aθ = 4.75 m/s² ρ = 153 m Style: Clean professional educational animation, 2D engineering/physics textbook style, white background, clear black labels, simple blue projectile trajectory, smooth motion, arrows for vectors, equations appearing step by step, camera zooming into important calculations, no unnecessary decorations. Make the animation easy for an engineering student to understand. Use accurate mathematical notation and keep all numerical values visible and readable.

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