in a stably stratified ocean, water that is displaced vertically from its equilibrium position will feel a restoring force that is proportional to the magnitude of the displacement. associated with this restoring force is a natural frequency of oscillation known as the brunt-vasiala frequency az where n is the brunt-vaisala frequency, g is the acceleration due to gravity, ρ is the density of the displaced water. derive this equation for n2. you could start by writing down the one-dimensional equation for motion for the vertical displacement of a water element displacement downward in a density stratified fluid and recall equations for simple harmonic motion.

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

the answer is c i did the same question so i know that's right!

we need to know the time interval during which the force acts. impulse of force = momentum change