Please, help!!
Three equal point charges of 4.00 nC lie 4.00 m apart on
a line. Calculate the magnitude and direction of the net
force on the middle charge.

The net force in the middle particle is zero, and it has no direction.

Explanation:

First, the force that one charge does in other charge is:

F = K*q1*q2/r^2

where k is a constant, q1 and q2 are the charges, and r is the distance between the charges.

If the force is positive, the force is repulsive (pushes away the charge) if the force is negative, is attractive.

Now, we have that the middle charge has a charge of 4.00nC, and in each side at a distance of 4m, has a charge of 4.00nC.

i will write qL as the charge in the left, qR as the charge in the right, and qM as the charge in the middle.

The force that the charge in the right does to the charge in the middle is:

Fr = (k*(4.00nC)^2)/4m^2 = k*(nC/m)^2

And is positive, so this is a repulsive force, this means that if the charge is at the right of the middle charge, then this force pushes the middle charge to the left.

For the left charge we have the same:

Fl = (k*(4.00nC)^2)/4m^2 = k*(nC/m)^2

But in this case, the force pushes the particle to the right, then this force, that is equal in magnitude to the previous force, pushes it in the opposite direction, then the total force in the middle particle is anulated. This is because the two external particles are "pushing" the middle particle with the same force and in the opposite direction.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be