The figure below shows a worker poling a boat-a very efficient mode of transportation-across a shallow lake. He pushes parallel to the length of the light pole, exerting on the bottom of the lake a force of 225 N. Assume the pole lies in the vertical plane containing the boat's keel. At one moment, the pole makes an angle of 30.0° with the vertical and the water exerts a horizontal drag force of 47.5 N on the boat, opposite to its forward velocity of magnitude 0.857 m/s. The mass of the boat including its cargo and the worker is 360 kg. (a) The water exerts a buoyant force vertically upward on the boat. Find the magnitude of this force. (b) Model the forces as constant over a short interval of time to find the velocity of the boat 0.400 s after the moment described. I got the answer for (a), 3.34e3 N. I calculate: [Force of gravity] = [mass][gravity] = [360kg][9.81m/s^2] = 3531.6 N [Normal force] = [Force exerted on bottom of the lake]*cos30 = 225N*cos30 = 194.9 N [Magnitude of the force] = [Force of gravity]-[Normal force] = 3531.6 N-194.9 N = 3337N But I couldn't find the answer for (b) What I got so far is to calculate the: (Net force) = (Horizontal drag force) + [(Force exerted on the bottom of the lake)*(cos 60)] = -47.5N + (225*cos 60) = 65.0N then find accerleration, a a = F/m = 65.0N/360kg = 0.180556m/s^2 then find v(0.400) v = at = 0.180556m/s^2 * 0.400s = 0.07222m/s then add both velocities 0.007222m/s + 0.857m/s = 0.9292m/s = 9.29e-1m/s and I don't understand why it's a wrong answer. Can you figure out why?