The ball rises to a height of approximately 1.9895 meters after the bounce.
To determine the height to which the ball rises after the bounce, we can make use of the principle of conservation of mechanical energy. In this case, we assume that there is no energy lost due to air resistance.
The initial potential energy of the ball can be calculated using the mass and height at which it is dropped. Since the height is not provided in the question, we'll assume it is denoted as 'h'.
Initial potential energy (before the bounce) = mgh
where
m = mass of the ball = 28.0 grams = 0.028 kg (converting grams to kilograms)
g = acceleration due to gravity = 9.8 m/s² (approximately)
The initial potential energy of the ball is given as 1.30 joules. Therefore, we have:
mgh = 1.30 joules
0.028 kg * 9.8 m/s² * h = 1.30 joules
0.2744 h = 1.30 joules
h ≈ 1.30 joules / 0.2744
h ≈ 4.7344 meters
So, the ball is dropped from a height of approximately 4.7344 meters.
Now, we'll determine the height to which the ball rises after the bounce, taking into account the 58% efficiency of the collision.
The efficiency of the collision is given as 58%, which means 58% of the initial kinetic energy is retained after the bounce.
Let the final height to which the ball rises after the bounce be denoted as 'H'. The final kinetic energy of the ball can be calculated using the equation:
Final kinetic energy (after the bounce) = (1 - 58%) * Initial kinetic energy
The initial kinetic energy of the ball is equal to the initial potential energy, which is 1.30 joules. Therefore:
Final kinetic energy (after the bounce) = (1 - 0.58) * 1.30 joules
Final kinetic energy (after the bounce) = 0.42 * 1.30 joules
Final kinetic energy (after the bounce) ≈ 0.546 joules
The final potential energy of the ball is given by:
Final potential energy (after the bounce) = mgh
0.028 kg * 9.8 m/s² * H = 0.546 joules
0.2744 H = 0.546 joules
H ≈ 0.546 joules / 0.2744
H ≈ 1.9895 meters
Therefore, the ball rises to a height of approximately 1.9895 meters after the bounce.