1. Velocity of the Projected Ball ($v_1$): Using $v = u - gt$ (upward motion):
$$v_1 = 40 - 10t$$
2. Velocity of the Dropped Ball ($v_2$): Using $v = u + gt$ where $u = 0$ (downward motion):
$$v_2 = 0 + 10t = 10t\lt strong\gt 3. Equating Magnitudes:\lt /strong\gt |v_1| = |v_2|$$
$$40 - 10t = 10t$$
(Note: Since we are looking for a point where both are still in motion, $v_1$ is still positive/upward).
$$40 = 20t \implies t = \frac{40}{20} = 2 \text{ s}$$
At $t = 2 \text{ s}$, the first ball has a velocity of $40 - 20 = 20 \text{ ms}^{-1}$ (upward), and the second ball has a velocity of $20 \text{ ms}^{-1}$ (downward). Their magnitudes are equal.