A golf ball strikes a hard, smooth floor at an angle of 38.8 and, as the drawing

Question

A golf ball strikes a hard, smooth floor at an angle of 38.8 and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0682 kg, and its speed is 53.4 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, and ignore the weight of the ball.)

I need to understand how to work this and how to apply the formulsa correctly

Thank you

Explanation / Answer

assuming that it's 27

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