A batter in baseball hits a pop-up straight up in the air from a height of 1.365 m. The ball rises to a height of 7.679 m above the ground. Ignoring air resistance, what is the speed of the baseball when the catcher gloves the ball 1.617 m above the ground?

To solve this problem, we can use the principles of physics, specifically the law of conservation of energy. This law states that the total energy of an isolated system remains constant—it is said to be conserved over time. In this case, we can consider the baseball as the isolated system. The total energy of the baseball is the sum of its kinetic energy (energy due to its motion) and potential energy (energy due to its height).

1. First, we need to calculate the initial speed of the baseball when it was hit. We know that the maximum height reached by the ball is 7.679 m and the initial height is 1.365 m. So, the height gained by the ball is 7.679 m - 1.365 m = 6.314 m.

2. We can use the formula for potential energy to find the initial speed. The potential energy at the maximum height is mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s^2), and h is the height. However, since we're looking for the speed (which is not dependent on the mass of the ball), we can ignore the mass in our calculations.

3. The potential energy at the maximum height is g*h = 9.8 m/s^2 * 6.314 m = 61.8772 m^2/s^2.

4. This potential energy is converted into kinetic energy as the ball falls. The kinetic energy of an object is given by the formula 1/2mv^2, where m is the mass and v is the velocity. Again, since we're looking for the speed (which is not dependent on the mass of the ball), we can ignore the mass in our calculations.

5. Setting the potential energy equal to the kinetic energy gives us 61.8772 m^2/s^2 = 1/2v^2. Solving for v gives us v = sqrt(261.8772 m^2/s^2) = 11.18 m/s. This is the speed of the ball at the maximum height.

6. Now, we need to find the speed of the ball when the catcher gloves it at a height of 1.617 m. The potential energy at this height is g*h = 9.8 m/s^2 * 1.617 m = 15.8466 m^2/s^2.

7. The total energy of the ball (the sum of its kinetic and potential energy) remains constant, so the kinetic energy of the ball at this height is the total energy minus the potential energy, or 61.8772 m^2/s^2 - 15.8466 m^2/s^2 = 46.0306 m^2/s^2.

8. Using the formula for kinetic energy, we can find the speed of the ball at this height: v = sqrt(2*46.0306 m^2/s^2) = 9.61 m/s.

So, the speed of the baseball when the catcher gloves the ball 1.617 m above the ground is approximately 9.61 m/s.