There is a ball that is thrown at a 50 degree angle traveling 47 centimeters a second. Every time the ball bounces, the ball regains 87% of it height. How far and high is the ball from it original start, x, after 10 seconds?

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We have a projectile motion problem with bouncing. A ball is thrown at a 50-degree angle with an initial velocity of 47 centimeters per second. Each time the ball bounces, it retains 87% of its height. We need to find the ball's horizontal distance and height after 10 seconds. Let's analyze the initial trajectory. We decompose the velocity into horizontal and vertical components. The horizontal component is 47 cosine 50 degrees, which equals 30.2 centimeters per second. The vertical component is 47 sine 50 degrees, which equals 36.0 centimeters per second. Using kinematic equations, the flight time is 0.073 seconds, maximum height is 6.6 centimeters, and horizontal range is 22.1 centimeters. Now let's understand bounce mechanics. When the ball hits the ground, the horizontal velocity remains constant at 30.2 centimeters per second. However, the vertical velocity changes. The ball retains 87% of its height, which means the new vertical velocity is the old velocity multiplied by the square root of 0.87, approximately 0.933. So after the first bounce, the vertical velocity becomes 33.6 centimeters per second upward. Let's calculate multiple bounces systematically. The first bounce occurs at 0.073 seconds at 22.1 centimeters. The second bounce happens at 0.141 seconds at 42.7 centimeters. Each subsequent bounce has decreasing height following the 87% pattern. The flight times get shorter as the vertical velocity decreases. Within 10 seconds, the ball will complete approximately 137 bounces, with the horizontal distance continuing to accumulate. Now let's find the exact position at 10 seconds. Since the horizontal velocity remains constant at 30.2 centimeters per second, the total horizontal distance is simply velocity times time: 30.2 times 10 equals 302 centimeters. After many bounces, the ball's height becomes negligible, so it's essentially bouncing on the ground. Therefore, at t equals 10 seconds, the ball is 302 centimeters horizontally from its starting point and at zero height on the ground.