A woman stands in a boat at rest on a calm lake. She throws a 10 kg anchor off the front of the boat, which has a mass of 1,000 kg. a). What is the initial momentum of the anchor-boat system? b). In which direction will the boat move after the anchor is thrown? c). How does the change in momentum of the anchor compare to the change in momentum of the boat? d). If the velocity of the anchor is 12 m/s, what is the velocity of the boat after the anchor is released?

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

use photo math tells u all the answers thats what i use