Answer:
The magnitude of impulse the floor applies to the ball is 3.05 kg.m/s.
Explanation:
Given;
mass of the rubber ball, m = 0.25 kg
height of drop, h₁ = 2 m
height of rebounds, h₂ = 1.8 m
Determine the initial velocity of the ball as it moves downwards;
[tex]v_i = \sqrt{2gh}\\\\ v_i = \sqrt{2*9.8*2}\\\\v_i = 6.26 \ m/s[/tex]this initial velocity is acting downwards = - 6.26 m/s
Determine the final velocity of the ball as it rebounds
[tex]v_f = \sqrt{2gh} \\\\v_f = \sqrt{2*9.8*1.8}\\\\ v_f = 5.94 \ m/s[/tex]this final velocity is acting upwards = 5.94 m/s
Impulse is given by;
J = mΔv
[tex]J = m(v_f-v_i)[/tex]
J = 0.25(5.94 - (6.26))
J = 0.25(5.94 + 6.26)
J = 3.05 kgm/s
Therefore, the magnitude of impulse the floor applies to the ball is 3.05 kg.m/s.
