Two students are passing a ball back and forth, allowing it to bounce once between them. If one student bounce-passes the ball from a height of 1.3 meters and it bounces 3.4 meters away from the student, where should the second student stand to catch the ball at a height of 1.1 meters? Round your answer to two decimal places, if necessary.
To find out where the second student should stand to catch the ball at a height of 1.1 meters, we can use the concept of conservation of energy.
The initial potential energy (PE) of the ball when it is thrown by the first student at a height of 1.3 meters converts into kinetic energy (KE) when the ball reaches the maximum height after bouncing. The ball then loses potential energy as it gains kinetic energy while falling back down to the second student's catch height of 1.1 meters.
First, let's calculate the ball's initial potential energy:
PE = m * g * h
PE = mass of the ball * acceleration due to gravity * height
PE = m * 9.8 m/s^2 * 1.3 meters
Next, we'll calculate the ball's final potential energy at the height of 1.1 meters after the bounce:
PE = m * g * h
PE = m * 9.8 m/s^2 * 1.1 meters
Since energy is conserved, the initial potential energy of the ball must equal its final potential energy:
m * 9.8 m/s^2 * 1.3 meters = m * 9.8 m/s^2 * 1.1 meters
Now, let's solve for the mass of the ball. We can see that the mass cancels out from both sides of the equation:
1.3 = 1.1
This shows that the position where the second student should stand to catch the ball at a height of 1.1 meters is the same as the position where the first student bounce-passed the ball, which is 3.4 meters away. So, the second student should stand 3.4 meters away from the first student.