Tranmission line equations when initial conditions are not zero
The initial conditions are that the 20 &Omega; line is charged to 1 MV, and at t=0 you connect a 30 &Omega; load.

The immediate effect is that the line and the load form a voltage divider, so the terminal voltage at the load end immediately becomes 600 kV. This -400 kV step is the initial disturbance that propagates back up the line toward the source.

Both the source impedance and the load impedance are 30 &Omega;, so the reflection coefficient is the same at both ends:

$$\Gamma = \frac{Z_T - Z_0}{Z_T + Z_0} = \frac{30 \Omega - 20 \Omega}{30 \Omega + 20 \Omega} = 0.2$$

At the source, the initial -400 kV step is reflected as a -400 &times; 0.2 = -80 kV step, bringing the line to 520 kV.

At the load, this step is reflected as an additional -80 &times; 0.2 = -16 kV step, bringing the line to 504 kV.

The next steps are 500.8 kV, 500.16 kV and 500.032 kV.

After a large number of steps, the line voltage converges on 500 kV, just as you'd expect. In the steady state, you have a simple voltage divider between the source impedance and the load impedance.

So now we know what happens, but how fast does it happen?

Knowing the capacitance allows us to also calculate the line's inductance from its characteristic impedance:

$$Z_0 = \sqrt{\frac{L}{C}}$$

so,

$$L = Z_0^2 C = 20^2 0.2nF/m = 80nH/m$$

Knowing both of these values allows us to calculate the velocity factor for a lossless line:

$$VF = \frac{1}{c \sqrt{L C}} = \frac{1}{3\cdot10^8 \sqrt{0.2 nF \cdot 80 nH}} = 0.8333$$

Those two formulas can be combined into a simpler formula that gives the same answer:

$$VF = \frac{1}{c Z_0 C}$$

Knowing this, plus the length of the line, allows you to calculate the time between the reflections.