Final answer:
The magnitude of the impulse experienced by the golf ball during the bounce is approximately 0.036 kg·m/s.
Step-by-step explanation:
To calculate the impulse experienced by the golf ball during the bounce, we need to use the principle of conservation of momentum. Before the bounce, the ball is dropped from a height of 3.2 m and its velocity just before it strikes the ground can be calculated using the equation v₁ = √(2gh₁), where v₁ is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h₁ is the initial height.
After the bounce, the ball reaches a height of 2.12 m. To calculate the velocity just after it leaves the ground, we can use the equation v₂ = √(2gh₂), where v₂ is the velocity and h₂ is the final height.
The impulse experienced by the ball during the bounce can be calculated using the equation I = mv₂ - mv₁, where I is the impulse, m is the mass of the ball (0.1285 kg), v₂ is the final velocity, and v₁ is the initial velocity.
Using the given values, the calculations would be as follows:
v₁ = √(2 * 9.8 * 3.2) ≈ 7.01 m/s
v₂ = √(2 * 9.8 * 2.12) ≈ 6.73 m/s
I = 0.1285 * 6.73 - 0.1285 * 7.01 ≈ -0.036 kg·m/s
The magnitude of the impulse experienced by the golf ball during the bounce is approximately 0.036 kg·m/s.