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A ball whose mass is 0.2 kg hits the floor with a speed of 4 m/s and rebounds upward with a speed of 3 m/s. if the ball was in contact with the floor for 0.5 ms (0.5multiply10-3 s, what was the average magnitude of the force exerted on the ball by the floor?

Respuesta :

Impulse = Ft = Δmv, where F is the average force, t is time, m is mass and v is velocity.

Δmv = m(v₂ - v₁) = 0.2 * (3 - (-4)) = 0.2 * 7 = 1.4

Solve for F.

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