In this video, I go over projectile motion along an inclined plane and show that the launch angle that gives the maximum range upwards or downwards is half the angle between the plane and the vertical. Even when firing at a target up an inclined plane, the angle that minimizes the energy used is also half the angle between the plane and the vertical. This result also matches for a flat plane, since the maximum distance is achieved for a launch angle of 45°, i.e. half the angle between the plane (0°) and the vertical (90°).
Notes - 3Speak - YouTube - Telegram - Problems Plus - https://mes.fm/math
#math #calculus #projectilemotion #vectors #vectorfunctions
Timestamps
- Problem 2: Projectile fired down inclined plane – 0:00
- Solution to (a): Position vector and parametric equations of the path of the projectile down an inclined plane – 3:16
- Interval for t from launch to hitting the ground – 13:55
- Solution to (b): Maximizing distance of a projectile fired down an inclined plane – 14:48
- Distance for the range downhill – 22:21
- Range is maximized when the derivative is equal to zero – 24:11
- Angle that maximizes range is half the angle between the plane and the vertical – 31:40
- Solution to (c): Maximizing distance of a projectile up an inclined plane – 34:58
- Replace our angle with angle θ with -θ in Parts (a) and (b) since the steps are the same, including the result – 39:08
- Solution to (d): Minimizing energy requires the same launch angle as in Part (c) up an inclined plane to hit a target – 43:56
- Energy is minimized when the derivative of the square of velocity is zero – 48:46
- Angle to minimize energy is half the angle between the plane and the vertical – 58:19
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