Q&A for How to Calculate Terminal Velocity

It will continue to fall at that speed without accelerating until it hits something.

B.A.S.E is an acronym for Buildings, Antennas, Spans, and Earth. BASE jumping is an extreme sport in which you jump off these various forms and deploy a small parachute.

Depends on the weight, shape, and density of the air. Hence, you have a way to calculate it, as it is not a constant.

You can't. You need more information to solve the equation.

Anything with mass requires infinite force to reach the speed of light, so even in a vacuum it is impossible to reach the speed of light. This is really complicated, as light is composed of photons, which are massless particles. Due to this law, terminal velocity in a vacuum would be 99.999...% of the speed of light. According to Einstein's Theory of Special Relativity, this also causes time travel into the future, because approaching the speed of light slows down time. If you could travel faster than light, terminal velocity would be completely nonexistent (you'd keep accelerating). Basically, terminal velocity in space is just an infinitely tiny bit slower than light.

Momentum(kgm/s) is equal to mass(kg) times velocity(m/s). 950kg * 40m/s = 38000kgm/s 950kg * 31m/s = 29450kgm/s 29450kgm/s - 38000kgm/s = -8550kgm/s

By the definition of a free fall, there is no air resistance. Since terminal velocity is reached when the force of air resistance = the force of gravity, this question is a trick question.

Terminal velocity refers to the equilibrium velocity as an object falls through a medium. The boat has no velocity relative to the most relevant medium, the water. If the boat were sinking, then it could obtain a terminal velocity, in which case the discussion above applies.

If a skydiver has reached TV, then he is no longer accelerating downward. If you know the time he has fallen since reaching TV and the distance fallen, then his T.V.=dist/time. This is due to the fact the acceleration component is zero (due to being at terminal velocity).

You would be adding mass and possibly altering the drag. You would need to recalculate accounting for a different "object." It is possible the secondary object was affected by a lower air density due to the wake of the first object. It is also possible that drag changes drastically as airflow between two objects is very complicated. See aerodynamics like F1 having to account for vortexes. There will surely be some energy transfer at impact, but it would still return to its TV given enough time still in the air.

v = the square root of ((2*m*g)/(ρ*A*C)) but m=ρ*v, so v=the square root of((2*ρ*v*g)/(ρ*A*C)), so v=the square root of((2*v*g)/(A*C)). Finally v=the square root of((8/3*r*g)/c), r-radius.

Terminal velocity is a velocity, and the SI unit of velocity is meters per second (m/s).

The drag coefficient depends on the shape of the falling object. For simple standard shapes you can look it up in a table, but if it is more complex, then you can find it with an experiment and the drag equation F = ρ.v^2.C.A where F is the drag force, ρ is the density of the fluid (air), C is the drag coefficient, and A is the projected area.

If given only the radius of an object, that means it is probably a sphere, for which the drag coefficient is 0.47. Assuming the density is of the sphere, not of the air, you can find its mass because you can calculate its volume from the radius. You also can find the projected area, which I believe would be half the surface area. The temperature should help somewhat in finding the density of air, although there are other factors that affect air density.

Terminal velocity is not dependent on initial velocity, but it is dependent on the force of gravity. You can calculate the terminal velocity using the acceleration due to gravity as well as other variables as stated in the equation.