The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long alnico magnet is 0.10 t. to produce the same field with a solenoid of the same size, carrying a current of 2.1 a , how many turns of wire would you need?

N = 3032  turns

Explanation:

The magnetic field produced by a solenoid is described by

        B = μ₀ n I

Where is the permittivity in a vacuum with a value of 4π 10⁻⁷ N /A²,  n  is the turn density and I the current

Let's apply this equation to the problem, the turn density is the number of turns per unit length, in this case it is the same magnet length

       L = 8 cm = 0.08 m

Let's calculate

      B = μ₀ N/L   I

      N = B L / μ₀ I

      N = 0.10 0.08 / (4π 10⁻⁷  2.1)

      N = 3,032 103 turns

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