For question 1 if you have your acceleration equal to -9.8m/s^2 then you can find how fast it is traveling when it hits the water. You would use the equation Vf=Vi+a(t). You are looking for your Vf. Vi is 0m/s. a is -9.8m/s^2. t is 3.87s. If you now plug in what you have and solve you would end up with Vf=-37.926. For how far did it drop use the equation, d=Vi(t)+1/2(a)(t)^2. plug in the Vi, a, and t and solve. Then you are left with the distance vertically being 73.38681m (don't forget sig figs). (When you solve for d it ends up being negative, but because you can't ever go backwards in distance it has to be positive)
