63 Physical Model for Terminal Velocity

After jumping, a skydiver begins gaining which increases the they experience. Eventually they will move fast enough that the air resistance is equal in size to their , but in opposite direction so they have no . This processes is illustrated by for a skydiver with 90 kg mass in the following image:

Free body diagrams showing the vertical forces of drag and gravity and resulting acceleration on a person at four times during a skydive from initial drop to terminal velocity. For all times the force of gravity is -888 N. The other example values are drag = zero, acceleration = -9.8 m/s/s; drag = 300 N, acceleration = -6.5 m/s/s; drag = 600 N, acceleration = -3.1 m/s/s, drag = 882 N, acceleration = 0.
Free body diagrams of a person with 90 kg mass during a skydive. The initial speed is zero, so drag force is zero. As speed increases, the drag force grows, eventually cancelling out the person’s weight. At that point acceleration is zero and terminal velocity is reached.

Dynamic Equilibrium

With a net force of zero the skydiver must be in , but they are not in because they are not static (motionless). Instead they are in , which means that they are moving, but the motion isn’t changing because all the forces are still balanced (net force is zero). This concept is summarized by , which tells us that  an object’s motion will not change unless it experiences a net force. Newton’s first law is sometimes called the Law of Inertia because is the name given to an object’s tendency to resist changes in motion.   applies to objects that are not moving and to objects that are already moving. Regarding the skydiver, we are applying Newton’s First Law to (back and forth, up and down), but it  also holds for the effect of on changes in rotational motion.  Changes in motion are known as accelerations and we will learn more about how net forces cause translational accelerations in upcoming chapters.

Reinforcement Exercises

Using above the statement of Newton’s first law as it applies to net forces and translational motion as an example, write out Newton’s 1st Law as it applies to torques and changes in rotational motion

Dependence of Terminal Velocity on Mass

We already know from our experimental work during the Unit 3 lab that increasing leads to increasing . We can now understand that this behavior occurs because greater mass leads to a greater and thus a greater speed is reached before the () is large enough to balance out the weight.

Everyday Examples

First-time skydivers are typically attached to an instructor (tandem skydiving). During a tandem skydive the bodies are stacked, so the shape and of the object don’t change much, but the does. As a consequence, the for tandem diving would be high enough to noticeably reduce the fall time and possibly be dangerous. Increasing the  to account for the extra mass is accomplished by deploying a small chute that trails behind the skydivers, as seen in the photo below.

Tandem skydivers with a small speed-limiting drag chute trailing behind. Image Credit: Fallschirm Tandemsprung bei Jochen Schweizer By Jochen Schweizer via Wikimedia Commons


A Physical Model for Terminal Velocity

When the skydiver has reached and remains in a state of , we know the size of the must be equal to the skydiver’s , but in the opposite direction. This concept will allow us to determine how the skydiver’s should affect terminal speed. We start be equating the air resistance with the weight:

    \begin{equation*} F_d = F_g \end{equation*}

Then we insert the formulas for and for of an object near Earth’s surface. We designate the in the resulting equation v_T because these two forces are only equal at terminal speed.

    \begin{equation*} \frac{1}{2}C_d \rho A_x v_T^2 = mg \end{equation*}

We then need to solve the above equation for the .

(1)   \begin{equation*} v_T = \sqrt{\frac{2mg}{C_d \rho A_x}} \end{equation*}

Everyday Examples

Let’s estimate the of the human body. We start with the previous equation:

    \begin{equation*} v_T = \sqrt{\frac{2mg}{C_d \rho A_x}} \end{equation*}

We need to know the , , of air, and of the human body.  Let’s use the authors 80 kg mass and the density of air near the Earth’s surface at standard pressure and temperature, \rho =  1.2 \,\bold{kg/m^3}.  Drag coefficient and cross sectional area depend on body orientation, so let’s assume a standard skydiving posture: flat, horizontal, with arms and legs spread. In this case the drag coefficient will likely be 0.4-1.3.  A reasonable value would be C_d = 1[2]. To approximate the cross-sectional area we can use the authors average width of 0.3 m and height of 1.5 m  for an area of  A_x = 0.3 \bold{m} \times 1.5 {m} =  0.45 \,\bold{m^2}

Inserting these values into our terminal speed equation we have:

    \begin{equation*} v_T = \sqrt{\frac{2(80\,\bold{kg})(9.8\,\bold{m/s^2})}{1(1.2 \,\bold{kg/m^3}) (0.45\,\bold{m^2})}}= 40 \,\bold{m/s} = 120 \bold{MPH} \end{equation*}

Reinforcement Exercises

You already have data on how the depends on . We acquired this data using coffee filters in the Unit 3 Lab. Looking back at that data, does that data support our physical model for terminal speed? [Hint: If our (fit equation) suggests that terminal speed depends on mass in the same way as the then yes, our data supports our physical model. Our physical model says that the terminal speed depends on the square root of the mass. Does your empirical fit equation support that result?]

Acceleration During a Skydive

We have now analyzed the skydive after was reached. Prior to this point the forces of drag and weight are not equal, therefore the skydiver is not in and speed will change over time. In order to analyze the early part of the skydive, or the motion of any object that is not in ,  we need to learn how to quantify changes in and/or direction.


Share This Book