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As people in the comments pointed out, the pressure exerted at the ground will go to 0(due to rocket thrust) very quickly if the atmosphere is present. To solve that problem exactly, you will need to solve Navier's stokes equation, which as you would probably know, is difficult because of the non-linearity and has to be done numerically. I can give you some idea about what will happen in a vacuum. Let's assume the rocket is using ion thrusters and ejecting gas it at a speed $v_{rel}$ with mass rate $dm/dt$ relative to the rocket. As the gas is not hindered by particles in the atmosphere and neglecting the collisions due to it's own fuel bouncing back from the ground, it will reach the ground with speed $v_{fin} = \sqrt{({v_{rel}+v_{rocket}})^2+2*g*h}$, assuming g to be constant at small heights. Now, the pressure caused due to the gas is $F/A$ which is $\frac{2v_{fin}dm}{A_{fin}dt}$, assuming elastic collisions. In this equation, the $A_{fin}$ will be the effective area that the fuel lands in. Assuming the area of the rocket exhaust circular with radius $r_{init}$ and it shoots the fuel with a divergence of angle $\theta$, you will get $A_{fin}=\pi (r_{init}+h\tan \theta)^2$. You can make the formula more accurate by relaxing more and more assumptions.