please make a kinemics video for physics 20 to resolve difficult problem.

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We're solving a classic projectile motion problem. A ball is thrown horizontally from a 50 meter cliff at 20 meters per second. We need to find how far from the base it lands. Remember: motion in the x and y directions are independent! Let's analyze the vertical motion first. Since the ball is thrown horizontally, the initial vertical velocity is zero. We use the equation delta y equals one half g t squared. Substituting our values: 50 equals one half times 9.8 times t squared. Solving for time, we get t equals 3.19 seconds. Now for horizontal motion. Horizontal velocity remains constant at 20 meters per second. Using delta x equals v x times t, we multiply 20 meters per second by 3.19 seconds to get 63.8 meters. The ball lands 63.8 meters from the cliff base! Let's recap the key concepts. Projectile motion splits into two independent parts: vertical motion with free fall using delta y equals one half g t squared, and horizontal motion with constant velocity using delta x equals v x times t. Always solve vertical motion first to find the time of flight! Here's a challenge problem! Now the ball is thrown at 30 degrees above horizontal with 20 meters per second. Use velocity components: v x equals v cosine theta, and v y equals v sine theta. Follow the same steps: find the components, calculate time using vertical motion, then find the horizontal range. Good luck!