How do I find the output voltage of a load?
If you saturate that transistor, then you will pull its collector low. It will be nearly ground potential. If the base current times the transistor gain is greater than the current that gets through that resistor, then its collector should be "pulled" within millivolts of ground.

An easy way to understand it, is to understand voltage dividers. Voltage dividers work on the principle that the sum of the voltages is zero. Another way to say it is, all of the voltage "drops" across the loads if you follow any given path from positive to negative supply. The amount that drops across a given load is determined by what proportion of the total resistance that part is. If half the resistance is in one component then half of the voltage will drop across that component.

In your circuit, some of the voltage will drop across that resistor, and the rest drops across the transistor. When the transistor is on, it effectively becomes lower resistance, because more current flows through it, so it becomes a smaller proportion of the total resistance, so less voltage difference will be present across the collector-emitter junction. This "pulls" the collector to be closer to the emitter potential, which is ground.

When the transistor is off, very little current leaks through the transistor, so it represents a huge resistance. The transistor becomes the majority of the resistance, so, most of the voltage difference will drop across it. This causes the collector to go high, because little of the drop will occur across the resistor, and a lot of drop occurs across the transistor.