A basketball of mass 0.23kg is thrown horizontally against a rigid vertical wall with a velocity of 20m/s. It rebounds with a velocity of 15m/s. Calculate the impulse of the force of the wall on the basketball.

To calculate the impulse of the force of the wall on the basketball, we will use the principle of conservation of momentum. The impulse is equal to the change in momentum of the basketball, which is given by the following equation:

Impulse = Change in Momentum

Momentum = Mass × Velocity

The initial momentum of the basketball can be calculated as:
Initial Momentum = Mass × Initial Velocity

The final momentum of the basketball can be calculated as:
Final Momentum = Mass × Final Velocity

The change in momentum can be calculated by subtracting the initial momentum from the final momentum:
Change in Momentum = Final Momentum - Initial Momentum

Let's plug in the given values and calculate the impulse:

Mass of the basketball (m) = 0.23 kg
Initial Velocity (u) = 20 m/s
Final Velocity (v) = -15 m/s (since the ball rebounds in the opposite direction)

Initial Momentum = Mass × Initial Velocity = 0.23 kg × 20 m/s = 4.6 kg·m/s

Final Momentum = Mass × Final Velocity = 0.23 kg × -15 m/s = -3.45 kg·m/s

Change in Momentum = Final Momentum - Initial Momentum = -3.45 kg·m/s - 4.6 kg·m/s = -8.05 kg·m/s

Therefore, the impulse of the force of the wall on the basketball is -8.05 kg·m/s. The negative sign indicates that the impulse acts in the opposite direction of the initial momentum.