Mechanics accleration question and kinetic energy I am really confused by this and I don't know why:
Take the Earth to be spinning at 100,000 m/s at the surface.
If I throw a ball on earth at 1m/s then its true velocity is 100,001 m/s
Therefore, Change in KE = 0.5mass( 100,001^2 - 100,000^2 ) = 0.5mass200,001 J
Obviously it doesn't take this much energy to throw a ball on earth so why is this. In fact if you alter the problem such that it was 10m/s and 11m/s then it takes even less energy. Why would the change in kinetic energy be affected by the initial velocity when you are always doing the same increment. And therefore, why in mechanics problems do we take kinetic energies and velcoities relative to earth?
 A: We need a reference frame to know the velocity of a particle and earth provides a good inertial reference frame for calculation.
Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one.
So if an experiment is short enough and happens in a small  region, the surface of Earth can indeed be assumed to an inertial frame of reference since the effects on the experiment's results are very, very tiny.
By the way there is nothing as such true velocity. Velocity is frame depended.
A: The work done by a force depends on the choice of the reference frame. Let's call $S$ the Earth frame and $S'$ the other one. Let $v$ be the relative velocity, in this case $v=100m/s$. Let $F$ be the force that accelerates your ball. For simplicity let's suppose this force is costant. Let $t$ be the time that you take to accelerate this ball. What happens in $S$? Well, the displacement of the ball is $s=\frac{1}{2}\frac{F}{m}t^2$. So the work done is $W=\frac{1}{2}\frac{F^2}{m}t^2$. While in $S'$ we have $s'=\frac{1}{2}\frac{F}{m}t^2+vt$. So the work $W'=\frac{1}{2}\frac{F^2}{m}t^2+Fvt$.
Now, we all know that $\Delta K=W$. So it sould be no sorprise that you find different variation of kinetic energy. This is consistent with our calculation. Infact in $S$, $\Delta K=\frac{p_f^2}{2m}$ and $p_f=Ft$ (impulse theorem, or if you want $p_f=m\frac{F}{m}t$). While in $S'$ we have $p'_i=mv$ and $p'_f=m(v+\frac{F}{m}t)$. So $(p'_i)^2=m^2v^2$ and $(p'_f)^2=m^2(v^2+\frac{F^2}{m^2}t^2+2v\frac{F}{m}t)$. This way $\Delta p^2=F^2t^2+2vmFt$. So $\Delta K=\frac{F^2t^2}{2m}+vFt$ and everything is ok.
