Final answer:
Using the work-energy principle, the final kinetic energy of the 3 kg ball is found to be 69 J, considering the initial kinetic energy and the work done by gravity during free fall from a height of 1.5 m. From this, the final velocity of the ball just before impact is calculated to be approximately 6.78 m/s.
Step-by-step explanation:
To determine the velocity of the ball just before it hits the ground, we can use the work-energy principle, which states that the work done by all forces acting on an object equals the change in kinetic energy of the object. In this case, the net force due to gravity does work on the ball and changes its kinetic energy. The work done by gravity (W) is equal to the force of gravity (F) multiplied by the distance (d) over which it acts in the direction of the force, which in this example is 1.5 m downwards.
If we let down be positive, W = Fd = 30 N × 1.5 m = 45 J. The initial kinetic energy (KE_initial) of the ball is calculated using its mass (m) and initial velocity (v_i), given by ½ m v_i^2. Substituting the given values, KE_initial = ½ × 3 kg × (4 m/s)^2 = 24 J. The final kinetic energy (KE_final) is the sum of the initial kinetic energy and the work done by gravity: KE_final = KE_initial + W = 24 J + 45 J = 69 J.
Finally, we can find the final velocity (v_f) by rearranging the kinetic energy formula to solve for v_f: KE_final = ½ m v_f^2. Isolating v_f, we get v_f = √(2 × KE_final / m). Plugging in the final kinetic energy and the mass of the ball, we find v_f = √(2 × 69 J / 3 kg) = √(46) ≈ 6.78 m/s. Hence, the velocity of the ball just before it hits the ground is approximately 6.78 m/s.