How come I can kick a football further when it's moving towards me? If a ball is moving towards me, I can kick it further than if I were to kick it if it was stationary. But surely if the ball is moving in the opposite direction, it should take more force to kick it the same distance as I am accounting for the initial movement. 
Does this have a theoretical explanation, or is it that I can perform better technique if it's moving?
 A: The first comment above hits the answer right on the head, but I'll give some more details.
The interaction between your foot and the soccer ball is inelastic - which shouldn't be surprising because all real-life interactions are essentially inelastic, but it's more elastic then a collision between say, a concrete block and your foot, because of the construction of the ball. The ball is essentially made to store elastic energy.
To study the energy of this system, let's call the work you can do on the ball (no matter if it's moving or not) $W$. So, if the ball is stationary, you give the ball energy $W$, some energy is stored in the ball elastically when it deforms, but when it returns to shape the energy is converted to kinetic. So the kinetic energy of the ball is $W$.
If the ball is moving towards you, with kinetic energy $K$, and then you kick it, what happens? Well first, you have to make sure you kick it hard enough for it to turn around, so roughly $K>W$ (you should actually do momentum conservation for this, but let's assume the collision is essentially elastic). The instant the ball is stopped, the elastic energy of the ball is $U=K+W$ - all of the energy has gone into deforming the ball. But when the ball returns to shape, all of the elastic energy goes into kinetic, and so the final kinetic energy of the ball is 
$$K_f=U=K+W$$
that is, more then the energy you could give it when you kicked it from a stationary position.
Compare this to kicking a concrete block, which does not have the ability to store energy elastically. If you stop the block, all the energy $K+W$ goes into heating/destroying the block - or actually, destroying your foot!
So there are certainly some effects I am ignoring, like the fact that the interaction doesn't happen instantaneously and there are some sources of energy which I am not keeping track of, but in essence this is what happens. 
A: If would certainly take more force and time (impulse force $F\Delta t$) to reverse the speed of the ball than to just accelerate it to the same final speed, since $m(V_{init}+V_{final}) \gt mV_{final}$.
And you can feel the difference in force (pain), if you kick a ball moving at you at high speed, provided that you kick it "the same way", or more precisely with the same speed, as you would kick a resting ball. 
Whether the reversed ball moves further or not, depends on whether you are willing to tolerate the pain and whether can maintain the speed of your foot over the duration of the kick.
If you can, the reversed ball should move further. To see it, you can model your foot as a moving wall. If the wall was stationary, the ball would have bounce at about the same speed as it hit the wall (assuming low losses). If the wall is moving, the speed of the moving wall should be added to the final speed of the ball (consider the reference frame associated with the moving wall). 
A: I will give a simpler explanation. A ball hitting a wall, reverses direction and if friction is small, it is elastically scattered, i.e. keeps its speed. Momentum is conserved  because the mass of the wall is very large . It is more complicated in real situations, see the answer here  for details.
So, if your leg, where the ball hits,  were a wall you need not move and the ball would be reflected with its incoming  speed. The extra energy you provide with the kick increases its speed. Of course you are not a wall, you balance through your contact to the earth,and because you will move some and there will be friction,it will not really be an elastic collision  , still your weight is  many times larger than the ball's, and you can take advantage of this initially favorable situation to reflect the ball with a higher speed than it came in.
A: A short answer to this: I will treat the event of the ball being kicked as an elastic collision between two bodies (the foot (with the body attached) moving with initial velocity $u_1$, and the ball approaching with negative velocity $u_2$. 
The problem of elastic collisions of 2 particles is a problem from various physics courses, I won't derive it here, but I'll use the result for $v2$, the velocity of the football after the collision occured, given in Wikipedia:
\begin{align}
v_{2}={\frac {u_{2}(m_{2}-m_{1})+2m_{1}u_{1}}{m_{1}+m_{2}}}
\end{align}
The mass of the foot is bigger than the mass of the football, which makes the quantity $(m_2 - m_1)$ negative. As $u_2$ is negative as well, the velocity $v_2$ will increase for bigger absolute values of $u_1$.
Of course this argument is just an approximation to reality, where the foot is connected to the leg (and the leg to the body and so on). An easy way to model this would be to increase the mass of the foot, taking into account that it is attached to a leg and to a body. 
A: If a ball is moving toward you it will exert a force on you.If you kick it in opposite direction of its its inital direction it will require more force for same distance in case it was moving away from you.You hit it.Now the ball will cover same distance with less force than first case as by now there is already a force acting on ball in that particula direction with the force of your kick added.
