Agolf ball strikes a hard, smooth floor at an angle of 39.8 ° and, as the drawing shows, rebounds at the same angle. the mass of the ball is 0.0285 kg, and its speed is 30.4 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.)

The magnitude of the impulse is 1.33 kg m/s

Explanation:

please look at the solution in the attached Word file

   I = 1.1 N s

Explanation:

The momentum is equal to the change of the moment

       I = Δp = m -m v₀

They indicate that the ball arrived at an angle of 39.8º, let's use trigonometry to find the velocity component

      sin 39.8 = / v

      cos39.8º = vₓ / v

      = V sin 39.8

      = 30.4 sin39.4

      = 19.3 m / s

Let's calculate the momentum on the y axis

      I =  m   - m

      I = 0.0285 19.3 - 0.285 (-19.3)  

      I = 1.1 N s

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y