Bullet hits Earth at speed $0.99c$ According to wikipedia, the speed of the neutrino is as close as we can get to the speed of light with an object with rest mass.
As per the comments, since neutrinos come in different energies, let's assume this speed to be 0.99c.
Now let's assume it is possible to speed up a bullet to this speed.
I was wondering how big of an impact this would have on Earth.
Let's assume the atmosphere will not burn the bullet, the bullet will still be intact when hitting Earth.
Let's take the bullet weight 4.2g.
Relativistic effects apply, so 

The first term is what we will use, kinetic energy, and we do not use the second, rest energy.
E=2.4*10^19 J
I do not know if I am right with this calculation.
Question:


*

*How big would the impact be?

 A: You cannot ignore the $-m_0c^2$(minus $m_0c^2$)term as it is ofcourse not negligible as it still has a $c^2$ to it. So hence it would yield a Kinetic energy for the bullet to be around: $2.298*10^{15} J$ of energy(considering rest mass of bullet, $m_0$ to be $4.2g=0.0042kg$ and speed of light, $c=299792458$$m/s$)
By comparison the bomb on Hiroshima energy of had around 15 Kilotons of TNT or about $6.3*10^{13} J$. Thus assuming that the bullet does reach the surface with that exact energy(which is highly inaccurate as it would almost instantly vaporize due to collisions with air molecules in the atmosphere) it would be equivalent to about 36 Hiroshima impact bombs. This ofcourse in reality (if hitting a planet with no atmosphere for which the bullet could indeed reach the surface) would cause widespread destruction and a large radiation shockwave, could even result in a very deep hole as the bullet is small and the energy is concentrated.  
A: The way of approaching these impact scenarios is to consider the sequence of events, not just the energy. 
In the frame of the bullet, what happens? It essentially runs into a wall of rapidly moving matter and explodes. In fact, this matter behaves as an intense heavy ion beam moving at $0.99c$: given that nuclear cross-sections are on the order of $10^{-28}$ m$^2$ and there are about $10^{25}$ atoms per cubic meter in air (at sea level) we should expect for any given atom in the bullet to hit an air molecule after a few kilometres. The path length will be longer in the upper atmosphere but the result still stands: somewhere in the atmosphere each atom in the bullet will hit an air atom. Had the atmosphere been thinner it would have reached the ground before this happens.
( One very simple way of modelling an impactor is Newton's penetration law: the impactor will penetrate to about the depth that it has displaced about the same mass as it has itself. This is a crude estimate, but it can help us guess where the projectile will have to stop. Since air density scales as $\rho(t)=\rho_0 e^{-z/H}$ we should expect the impactor with mass $m$ and area $A$ to stop when $m=\rho_0 A \int_z^\infty e^{-u/H} du$ or $z=-H \log(m/\rho_0 A H)$. For $A\approx 1$ cm$^2$, $m=4.2$ g, $H=8500$ m and $\rho_0=1$ kg/m$^3$ I get $z\approx 45$ km. )
Note that the electrons of atoms are more spread out and doubtless will have a far higher interaction cross section with each other, but since they are much lighter the heating from them scattering will likely be less important than the nuclear process.
Now, what happens when two nuclei moving at relative speed $0.99c$ ($\gamma\approx 7$) hit each other? This is basically what people do in relativistic heavy ion colliders, although their beams have gamma factors one or two orders of magnitude above this. The result is basically that the nuclei scatter into a shower of fragments, mostly lighter nuclei with the heavy nuclei barrelling on for a while. Since these fragments are charged they are stopped over a few cm to meter distance and transfer their energy to the surrounding air or bullet. 
So, at this point the bullet disintegrates. It does that fairly rapidly but the fireball will be very elongated since everything has enormous downward momentum. Basically there is relativistic beaming focusing it into a cone with an angle of 10-20 degrees. But again, the fireball will not penetrate many times its own mass downwards, so it remains high altitude. 
The total energy is as Tausif pointed out equivalent to a big nuclear blast ($\approx 639$ kt). However, it happens in the upper atmosphere so there is relatively little radiation and heat that hits the ground. There will be a shock-wave, but also plenty of distance to disperse it over. It is somewhat similar to a meteor airburst, slightly above the Chelyabinsk event but below Tunguska. The fireball will be far hotter but I think the altitude is more than enough to prevent surface fires or damage. The main special effect is likely the electromagnetic pulse from creating a plasma fireball up there.
If there is no atmosphere, then the impact explosion occurs on (or rather, below) the ground. Again the penetration law tells us that it will not go very deep - the 4.2 g bullet may have a relativistic mass 7 times larger, but this still only means it can traverse a few decimetres before having its momentum dispersed, and even then much of it is downward. The energy released produces an explosion not dissimilar to an underground nuclear explosion. Overlying material is ejected rather efficiently, producing a larger crater than a direct surface impact/explosion of the same yield.
