To do this you would probably first want to express the position of the deck as a function of time, then solve for that equal to the level of the wharf.
Basically you let y = height of the deck of the ship compared to the wharf,
then y'' + ay = 0 (SHM) for some a.
then you somehow solve it.. to get y = Acos(wt + c) + K where A, w and c are constants to be determined (A is called the amplitude, w is called teh angular frequency and c is the phase shift I think. K is just the center of motion, in this case it's the average of -2.4 and 0.6 which is -0.9) , . this you can do from the given conditions.
ofcourse A = half the distance between high and low positions.. so that gives 1.5
Now, w will depend on the units you measure time in. But basically it's the number such that when multiplied by the time taken to complete one complete cycle gives 2*pi. Here, suppose you use mins are your unit, then it takes 365 mins to complete HALF a cycle, so it takes 730 mins to complete 1 cycle, and therefore w = 2*pi/730 = Pi/365.
Finally c depends on when you consider as your time zero. If you take 8:30 am to be zero, then you know that at t=0, the deck should be at it's lowest point, or
cos(w*0 + c) = -1, solving this gives C = Pi.
So tying this altogether, one has:
y(t) = 1.5*cos(Pi/365 * t + Pi) - 0.9
Now one solve for t in y(t) = 0 (the formal interpretation of deck being level with th wharf)
that gives 0.6 = cos(Pi/365 * t + Pi)
t = [arccos(0.6) - Pi]*365/pi + 2n*pi, [-arccos(0.6) - Pi]*365/pi + 2n*pi
are solutions, of which the smallest one is 257.26
Finally remember to add this to time zero to give final answer 12:47:15
