# Question Video: Calculting the Terminal Velocity of a Falling Object

A skydiver with a mass of 80.0 kg falls through air that has a uniform density of 1.23 kg/m³. The skydiver has a surface area of 0.140 m² and a drag coefficient of 0.690. What is the skydiver’s terminal velocity?

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### Video Transcript

A skydiver with a mass of 80.0 kilograms falls through air that has a uniform density of 1.23 kilograms per cubic meter. The skydiver has a surface area of 0.140 meters squared and a drag coefficient of 0.690. What is the skydiver’s terminal velocity?

We can call this terminal velocity 𝑣 sub 𝑇. And we’ll also write down the skydiver’s mass, cross-sectional area, drag coefficient, and the density of the air the skydiver falls through. As the skydiver falls down to Earth, there’s a gravitational force pulling the skydiver in that direction. Along with the gravitational force, there’s the drag force resisting the skydiver’s fall. If we recall that drag force is proportional to the square of an object’s speed, at the start of their fall, the skydiver has very little drag force. But as they fall faster and faster, this force increases. And eventually, we get to a point where the drag force is equal to the gravitational force on the skydiver. At this point, they are no longer speeding up. They’ve reached terminal velocity.

We can rewrite 𝐹 sub 𝐷 in terms of the parameters it relies on. And 𝐹 sub 𝑔, we know, is equal to the mass of the skydiver times 𝑔, where 𝑔, the acceleration due to gravity, is 9.8 meters per second squared. So one-half the skydiver’s area times the air density times the drag coefficient times the terminal velocity squared is equal to the skydiver’s mass times 𝑔. Or 𝑣 sub 𝑇, the skydiver’s terminal velocity, is the square root of two 𝑚𝑔 divided by 𝐴, the skydiver’s area, times the density, 𝜚, multiplied by the drag coefficient. When we plug in for all these values and solve for 𝑣 sub 𝑇, we find it’s equal to 115 meters per second. That’s the maximum speed that this skydiver will achieve.