Simplify the circuit by replacing the b-e junction with a diode. What is the voltage at the cathode (base)? It is 10V - 10V = 0V. So what is the voltage at the anode (emitter)? It will be a diode drop higher at about 0.6V. When the transistor turns on it in effect connect the emitter and collector internally and thus the 0.6V will appear on the collector. So Vout will start charging from -10V to (0V + 0.6V) with an initial current of 9.4mA (10-0.6V)/1k, that tapers off according to the RC time constant.