When the switch is closed for long time, the capacitor acts as open circuit ofr DC current, the circuit will be as shown below
Applying KCL at the node
Current through 1Ω
Voltage across 1Ω
Voltage across the open capacitor
When the switch is open at t=0
Applying KCL
i = i1 + i2
Using partial fraction method
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Taking inverse laplace transform
Time constant
At t=5τ