Throwing an object with air resistance say I were to throw a 1 kg ball vertically upwards 100 m/s, with g=10 and also air resistance $0.01v^2$ . By considering the forces I can determine that the speed at which it lands is less than the initial speed. But does this apply to all cases?
What I mean is that because for the downward motion the air resistance now acts upwards so the net acceleration downwards is smaller... could this potentially give the ball more time to stay in the air which would be enough to allow its speed to surpass its launch speed? I have a feeling that this is not true because it would violate conservation of energy of something but I am not sure about this and would appreciate if someone could clarify this for me.
So is it possible for the ball, on landing to the point of projection, to be faster that it was when it was launched?
 A: The ball will always return to the ground at a lower speed than it was launched. This is simply because it must lose energy to air resistance during its flight.
The ball starts off with a kinetic energy $E_0$. Some of this energy is "spent" in lifting the ball (i.e. it is converted into potential energy) and some of it is spent on moving the air out of the way (i.e. air resistance - it is converted to kinetic energy in the air molecules it pushes away). At the top, the ball stops and now has $E_p < E_0$.
It starts back down again, converting $E_p$ back into kinetic energy, but also losing some more energy as it pushes the air away again. It finally arrives back where it started with $E_f < E_p < E_0$. So it is moving slower than when it left.
Note also, that if it is launched high enough, the air resistance on the way down (which increases as its speed increases) will eventually balance the acceleration due to gravity and it will reach terminal velocity. So not only is the final speed at the bottom always lower than the launch speed, it tends towards a maximum value. Beyond a certain launch speed, it always lands at the same speed.
A: At best, without air drag, your projectile will be traveling at the same speed at any given height, regardless of the direction. So if you throw it upward at $v_0=100$ m/s at a height of $0$ m, without air drag it will be moving at a speed of $100$ m/s when it falls back into your hand.
If we consider the downward motion of the projectile, we can use Newton's 2nd Law to obtain the equation:

$m\frac{dv}{dt}=mg-cv^2$, 

which is a nonlinear first order differential equation. Solving this we obtain the solution: 

$v(t) = v_T tanh(\frac{gt}{v_T})$, $v_T=\sqrt{\frac{mg}{c}}$

where ($c = 0.01$, the coefficient of your air drag). Solving $v_T$ we obtain $316.2$ m/s, giving us the fastest possible speed that the ball will move in the downward direction.
Integrating, and assuming that, at best (absence of air drag), the projectile can achieve a maximum height of $500$ m, you can obtain the equation of position:

$y(t)=y_{max}+\frac{m}{2c}\ln{(1-\tanh{(\frac{gt}{v_T}})^2})$,

which gives a time to fall from max height of approximately $t=7$ s. At this time, the $v(7)\approx70$ m/s. In fact, you will notice that the amount of time spent in the air is actually less than the "ideal" (no air drag) time in the air of $10$ s. It's actually an optimistic estimate, considering that in the presence of air drag, the ball will not actually make it to the same height it would in the absence of air drag.
The fastest possible downward speed it can have is $v_T=316.2$ m/s, and in the case of an initial speed of $v_0=100$ m/s, it will have reduced its speed to approximately $70$ m/s. There is no case where the ball could be traveling faster on the way down if there are damping forces present.
