System response with initial condition
The transfer function corresponds to a first order system, you can either assume that it's an RL or RC network with input voltage.

For unit step input of 8v, 

Y(s) =  800/s.(0.4s+1) = 250 /s.(s+2.5).

By partial fractions and taking inverse laplace,

y(t) = 800 (1 - exp(-2.5.t) ).

At steady steady state y(t) =800 V.

Now, this can thought as an RC circuit with 800 volts input, so that at steady state the capacitor is charged to 800 volts.

Once at t = to, say you short the source ( Vin = 0 ),the capacitor discharges through the resistor R .

We already know from equation of y(t) that the time constant (RC) = 1/2.5 = 0.4.

So,the capacitor discharges as follows:

y(t) = 800 exp(-2.5t).

At t = 0.1 (after source is shorted);

y(0.1) = 800 x 0.7788 = 623.04 volts.