Motion of the First Ball (Dropped): Initial velocity $u_1 = 0$. Time taken to meet $t_1 = 18$ s. Acceleration $a = g = 10$ ms⁻². Using the kinematic equation for displacement $h = ut + \frac{1}{2}gt^2$ : $h = 0 \times 18 + \frac{1}{2} \times 10 \times (18)^2$ $h = 5 \times 324 = 1620 \text{ m}$

Motion of the Second Ball (Thrown): It is thrown 6 seconds later, so its time of flight is $t_2 = 18 - 6 = 12$ s. Initial velocity $u_2 = v$. It covers the same distance $h = 1620$ m. Applying the displacement equation: $1620 = v(12) + \frac{1}{2} \times 10 \times (12)^2$ $1620 = 12v + 5 \times 144$ $1620 = 12v + 720$ $12v = 1620 - 720 = 900$ $v = \frac{900}{12} = 75 \text{ ms}^{-1}$