The question is asking what v_final is, given that v_initial is at 300 feet. and v_initial is at 0 feet. We know there will be a constant downward acceleration of 32.15 ft/s^2. Use the following equation: v_final^2 = v_initial^2 + 2ah v_final^2 = (160 ft/s)^2 + 2(-32.15 ft/s^2)(300 ft) = 6310 ft^2/s^2 v_final = (6310 ft^2/s^2)^1/2 = 79.4 ft/s.
