<span>vf^2 = vi^2 + 2*a*d --- vf = velocity final vi = velocity initial a = acceleration d = distance --- since the airplane is decelerating to zero, vf = 0 --- 0 = 55*55 + 2*(-2.5)*d d = (-55*55)/(2*(-2.5)) d = 605 meters
In this scenario it is easier to take a person to the water-pool than to transport the people in the air, as the person's strength is increased by water upwards: