A 1.2 kg ball drops vertically onto a floor, hitting with a speed of 20 m/s. it rebounds with an initial speed of 5.6 m/s. (a) what impulse acts on the ball during the contact

The force acting on the ball is the same than the ball acts on the floor at the moment of hitting but in the reverse direction. So the ball produces a force downwards and the floor makes the same force in value but upwards. Then the value of the force that impulses the ball upward is: F = m * a where F is force or impulse, m is mass of the ball and a is acceleration (9.8 m/s2) F = 1.2 kg * 9.8 m/s2 = 11.76 kg*m/s2 = 11.76 Newtons (upwards)