If you willingly fall on your turn does it cost movement?
As a DM I'm thinking about how to handle things such as creatures who are immune to being prone, diving into the water from a height, or a flying creature diving off a cliff and taking a fall before taking flight. There are many circumstances where you can fall without taking damage or being knocked prone afterward, and I'm not sure how to determine movement following the dive.
For example, my character walks up to a ledge above a 10 foot drop to the ground below. After taking 5 feet of movement to step off the ledge, I allow myself to fall. When I hit the ground have I used 15 feet of movement or just 5?
Best Answer
Here are the rules on falling from the Basic Rules (or PHB p. 183):
The fall itself wouldn't cost movement (it's "forced movement", except in this case the thing forcibly moving you is gravity). However, if you take damage and thus land prone, it will cost movement equal to half your speed to get up afterwards:
Assuming you have a speed is 30 feet, if you spend 5 feet of movement to step off a ledge, you will take 1d6 bludgeoning damage and land prone. Then it will cost another 15 feet of movement (half your speed) to stand up and continue moving; alternately, you could remain prone and continue moving, but you'd only be able to move 12.5 feet. (Your DM may have you round down to 10 feet if you're playing on a grid map.)
However, if you are for some reason able to land without falling prone or taking damage, then you could move 5 feet and fall off the ledge, land on the ground below, then continue moving as normal.