Physics
posted by Tom Atlus .
A golf ball strikes a hard, smooth floor at an angle of θ = 29.6° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.041 kg, and its speed is 45 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor.)

first find the component of velocity perpendicular to the wall,that is the componnet of veloicty that will be refersed.
The change in velocity will be have twice this component...the component parallel does not change.
impulse=mass*2*velocityperpendicular. direction is away from the wall.
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