Buoyancy Force = (dV)g
m = d’V 
acceleration = net Force/m => dVg/d’V = > d*g/d’ = 10/0.8 = 12.5 m/s^2
net accln.  = 12.5 – 10 = 2.5 m/s^2.
 
depth it gets sinked down:
=>
(2g*2) = 0^2 + 2*g*s
s = 8 meter.
 
						

							
This question has three parts, viz.,
	
Speed of ball just before entering water
	
Acceleration of ball in water due to buoyancy of water
	
Depth attained by ball due to changed acceleration.
 
1.         Speed of ball just before entering water
                u = 0 m/sec
                v = ? m/sec
                s = 2 m
                Since v2-u2=2gs
                Therefore,
                v = 2√g m/sec
 
	
Acceleration of ball in water due to buoyancy of water
Density(d) = Mass(m)/Volume(V)
Therefore Mass of ball(m) = 0.8V Kg
Density of water (at 4°C) = 1g/mL or 1000 Kg/m3
            Downward force exerted by ball
            F = mg
Or       Fb = 0.8Vg (Downwards direction)
            Since ball will displace equal volume of water
Therefore Force due to buoyancy of water
            Fw = 1.0Vg (Upward direction)
Therefore Net force
            Fb – Fw = 0.8Vg – 1.0Vg = -0.2Vg (upward direction)
However,
            F = ma or a = F/m
            Where m is mass of the ball
Therefore,
            a = -0.2Vg/0.8V = -g/4 m/sec2
 
	
Depth attained by ball due to changed acceleration:
Initial velocity of ball (u) = 2√g m/sec
Final velocity of ball (v) = 0 m/sec
g = -g/4 m/sec2.
Hence
            (0)2-(2√g)2 = 2(-g/4) x s
Solving this gives us
            s = 8 m