A ball of 0.1 kg mass is dropped on a hard floor from a height of 0.45 m and rises to a height of 0.20 m. If it was in touch with the floor for 0.1 s, the net force it applied on the floor while bouncing is: (take the gravitational acceleration g = 10 m s⁻²)

Speed just before impact: v₁ = √(2gh₁) = √(2 × 10 × 0.45) = √9 = 3 m/s (downward). Speed just after impact: v₂ = √(2gh₂) = √(2 × 10 × 0.20) = √4 = 2 m/s (upward). Change in momentum Δp = m(v₂ − (−v₁)) = 0.1 × (2 + 3) = 0.5 kg m/s (upward). Average net force on the ball from the floor = Δp / Δt = 0.5 / 0.1 = 5 N upward. By Newton's third law, the net force the ball applies on the floor is 5.0 N.