A 600 g model rocket is on a cart that is rolling to the right at a speed of 2.5 m/s. The rocket engine, when it is fired, exerts a 8.0 N vertical thrust on the rocket. Your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point.

At what horizontal distance left of the loop should you launch?

8.66 m

Explanation:

600g = 0.6 kg

Let g = 10m/s2. Gravity acting on the rocket model would have a magnitude of:

mg = 0.6*10 = 6 N

When the thrust of 8N acting on the rocket, the net force would be F = 8 - 6 = 2N vertically upward. So the net acceleration according to Newton 2nd law would be

a = F/m = 2 / 0.6 = 3.33 m/s2

We can calculate the time t it would take to reach h = 20m vertically at this acceleration:

Assume that the horizontal speed v = 2.5m/s stays constant, then the horizontal distance from where it starts thrusting up to the loop is

2.5 * 3.46 = 8.66 m

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).