difficult physics problem

1. (a) Consider the vertical motion, use equation of motion s = ut + (1/2)a.t^2
with s = -10 m, u = v.sin(45), a = -g(= -10 m/s2), t = ?
hence, -10 = v.t.sin(45) + (1/2)(-10).t^2
Consider the horizontal motion,
50 = [v.cos(45)].t
solving the two equations for t and v gives t = 3.464 s, v = 20.41 m/s

(b) using conservation of energy,
(1/2)m(20.41)^2 + mg.(10) = (1/2)m.v^2
where m is the mass of the stone, and v is itsfinal speed
hence, v = 24.83 m/s
The direction is given by arc-cos[20.41cos(45)/28.83] degrees below the horizontal.

2. weight of sea water displacement = 1100.(1-x)g N, where g is the acceleration due to gravity, taken to be 10 m/s2
hence, upthrust = 1100(1-x)g N
Therefore, for the piece of wood to float,
1100(1-x)g = 880g
solve for x gives x = 0.2 m

3. (a) Thermal process in which the temperature remains constant.
(b) Using ideal gas law, PV/T = constant
(Po + d.g.h).V/(273+2) = (Po).(2V)/(273+25)
where Po is the atmospheric pressure, V is the original volume of the bubble, d is the desnity of water, h is the depth of the pond, g is the acceleration due to gravity.
Given that Po = 1.01x10^5 N/m2, d = 1000 kg/m3, g = 10 m/s2
solve for h gives h = 8.52 m

4. (a) Consider the vertical motion, use equation s = ut + (1/2)at^2
with u = 0 m/s, s = -50.sin(30) m, a = -g(= - 10 m/s2), t = ?
hence, -50.sin(30) = (1/2).(-10)t^2
solve for t gives t = 2.236 s

(b) Initial velocity = distance/time = 50.cos(30)/2.236 m/s = 19.37 m/s

(c) use conservation of energy,
(1/2)m.(19.37)^2 + mg(50.sin(30)) = (1/2)m.v^2
where m is the mass of the stone
solve for v