A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the planes x = d and x = -d. The y- and z-dimensions of the slab are very large compared to d and may be treated as essentially infinite. The slab has a uniform positive charge density p. Required:
a. Explain why the electric field due to the slab is zero at the center of the slab (x=0).
b. Using Gauss's law, find the magnitude of the electric field due to the slab at the points 0<= x <= d. Express your answer in terms of the variables p, x, d, and e_0.
d. Using Gauss's law, find the magnitude of the electric field due to the slab at the points x>= d. Express your answer in terms of the variables p, x, d, and e_0.
e. What is the direction of the electric field due to the slab at the pints x >= d.

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

okay i have so many questions. but the big one here is: um wt..f is this mess? ?