A 0.250-kg rubber ball is dropped from a height of 2.00 m. It hits the floor and rebounds to a height of 1.80 m. What is the magnitude of impulse the floor applies to the ball

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;

this initial velocity is acting downwards = - 6.26 m/s

Determine the final velocity of the ball as it rebounds

this final velocity is acting upwards = 5.94 m/s

Impulse is given by;

J = mΔv

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.

not really sure but thinking 1.54 mm/us

50 inches should be the resultant vector