Free Fall and Terminal Velocity

Terminal Velocity Formula

Terminal velocity is the maximum speed that any object can obtain when falling under the effect of gravity through a fluid, usually the atmosphere. It depends on several factors including the mass of the object, the cross-sectional area of the object, and the drag coefficient of the object.

It is described by the following equation.

$$V_t = \sqrt {\frac {2mg} {\rho AC_d}}$$

\(V_t\) is the final terminal velocity. \(m\) is the mass of the object. \(g\) is the acceleration due to gravity. \(\rho\) is the density of the fluid. \(A\) is the cross-sectional area of the object. \(C_d\) is the drag coefficient.

Terminal Velocity and Air Resistance

We mentioned that several factors can alter the terminal velocity experienced by an object. These factors can result in some interesting results!

Solving Problems involving Terminal Velocity

Now that we have gone through the different components that factor into the terminal velocity of an object, let's look at some practical exercises to test our understanding.

A skydiver is diving through the air at \(14,000\) feet and casually wonders how fast he is traveling. He knows his mass is \(80\) kilograms and his drag coefficient is only \(0.5\), thanks to his new skydiving suit. At his altitude and air temperature, the density of the air is \(1.2\) kilograms per cubic meter. Finally, he knows that diving towards the ground presents the smallest cross-sectional area at about \(0.5\) square meters. How fast is he falling through the air, how long will it take him to hit the ground, and can you solve the equation before he reaches the earth?

All we have to do is enter all the variables we are given into our equation for terminal velocity.

$$V_t = \sqrt {\frac {2mg} {\rho AC_d}}$$

$$V_t = \sqrt {\frac {2*(80kg)*(9.8\frac m {s^2})} {1.2\frac {kg} {m^3} * (0.5 m^2) * (0.5)}}$$

$$V_t = \sqrt {\frac {1568 \frac {kg m}{s^2}}{0.3 \frac {kg} m}}$$

$$V_t = \sqrt {5227 \frac {m^2}{s^2}}$$

$$V_t = 72 \frac m s$$

So our skydiver is falling at about \(72\) meters per second, which is about \(161\) miles per hour, or \(259\) kilometers per hour. Since he has reached terminal velocity, this will be his constant velocity until another force acts on him to change his velocity. In this case, that would be the force of the earth on his body. Luckily, we can find out how much time he has in which to pull his parachute before that happens.

$$Distance = Velocity * Time$$

$$14000 ft = 14000*0.305 m = 4270 m$$

$$4270 m = 72 \frac m s * t$$

$$t = 59.31 s$$

Our skydiver has just under a minute to fall before he reaches an abrupt conclusion to his flight!

However, he doesn't think that's quite enough time to enjoy the sensation of falling through the clouds, so he decides to change his position from diving to spread-eagled. He wants to fall for a full \(90\) seconds. Is the increase in cross-sectional area enough to give him \(30\) extra seconds of flight?

If he wants to take \(90\) seconds to cover \(4270\) meters of distance, he needs to fall slower.

$$4270m = V_{ff} * 90s$$

$$V_{ff} = 47 \frac m s$$

His new speed needs to be \(47 \frac m s\) if he wants a full 90 seconds of flight. What cross-sectional area will guarantee him that speed, with all other variables being equal?

$$V_t = \sqrt {\frac {2mg} {\rho AC_d}}$$

$$47 \frac m s = \sqrt {\frac {2*80kg*9.8 \frac m {s^2}}{1.2 \frac {kg}{m^3} * A * (0.5)}}$$

$$2209 \frac {m^2}{s^2} = \frac {1568 \frac {kgm}{s^2}}{0.6 \frac {kg}{m^3} * A}$$

$$2209 \frac {m^2}{s^2} * A = 2613 \frac {m^4}{s^2}$$

$$A = 1.18 m^2$$

In order to extend his flight as much as he wants, the skydiver needs to increase his cross-sectional area from \(0.5 m^2\) to \(1.18 m^2\). If he extends his arms and legs as wide as possible, it might be possible!

Difference between Free Fall and Terminal Velocity

Whereas terminal velocity is defined as the point where the resistive force of a fluid is equivalent to the accelerating force of gravity pulling an object down, free fall is defined as any situation where the only significant force acting upon an object is gravity.

Free Fall Formula

Given that gravity is the only force acting on an object in free fall, the equation for free fall is far simpler than that for terminal velocity.

$$V_{ff}=gt+V_0$$

\(V_{ff}\) is the final free fall velocity. \(g\) is the acceleration due to gravity. \(t\) is the time from the start of the fall. \(V_0\) is the initial velocity. We can see the differences between these states by examining and comparing free body diagrams.

Terminal Velocity Free Body Diagram

The forces on the skydiver are equal, with the air creating enough drag to perfectly cancel out the accelerating pull of gravity. This causes their velocity to stabilize, at least until conditions change.

A skydiver falling at terminal velocity experiences a downward force due to gravity, and an equal upward drag force due to air resistance. These equal forces result in no acceleration. StudySmarter Originals

Free Fall Free Body Diagram

However, for an object in free fall, there is no significant restorative force working against gravity. The acceleration due to gravity is unhindered, and so the velocity of the object continues to increase.

A skydiver above the atmosphere experiences free fall and does not have any force on his body besides gravity. Thus, his velocity continues to increase. StudySmarter Originals

Maximum Velocity for Terminal Velocity and Free Fall

The maximum speed achievable by an object in terminal velocity is dependent on the factors described above, but for the average human skydiver their terminal velocity will reach up to \(200 mph\), or \(320 km/h\). The maximum velocity achievable during free fall is unbounded. Without any resistive force acting against the acceleration of gravity, the velocity of an object in free fall will not stop increasing. Felix Baumgartner jumped from the edge of space in 2012 and was able to reach estimated speeds over \(843.6 mph\), or \(1,357.6 km/h\), before he reached enough atmosphere to experience a resistive force!

Free Fall Situations

This restriction on free fall, where gravity can be the only significant force acting on an object, narrows down the number of situations that are truly categorized as free fall.

For slow speeds of motion through the air, air resistance is negligible. This means that situations like throwing a ball in the air or jumping off the ground are free fall situations. Air resistance can be ignored, and the only significant force acting on those objects is the acceleration due to gravity.

Objects in space are also often in free fall. If there are no external propulsion systems engaged, the only force acting on spacecraft and astronauts is that of gravity.