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 a bouncing ball. The 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 position after 10 seconds.
First, we need to break down the initial velocity into horizontal and vertical components. The horizontal component is 47 times cosine of 50 degrees, which equals 30.21 centimeters per second. The vertical component is 47 times sine of 50 degrees, which equals 36 centimeters per second. We also need the coefficient of restitution, which is the square root of 0.87, approximately 0.933.
Now let's analyze the bouncing simulation. Each bounce segment takes time T equals 2 times vertical velocity divided by gravity. The total time for all bounces is approximately 1.091 seconds. After this time, the ball stops bouncing and moves horizontally along the ground for the remaining 8.909 seconds.
Now let's calculate the horizontal distance. During the bouncing phase from 0 to 1.091 seconds, the ball travels 32.95 centimeters. After bouncing stops, the ball moves horizontally for 8.909 seconds, covering an additional 269.20 centimeters. The total horizontal distance is 302.15 centimeters from the starting point.
Here is our final answer. After 10 seconds, the ball is located 302.15 centimeters horizontally from its starting point and at a height of 0 centimeters, meaning it's on the ground. The ball completed all its bounces within the first 1.091 seconds and then moved horizontally along the ground for the remaining time.