If an object of mass m falls in a vacuum (in the earth’s gravitational field) it accelerates with acceleration g, where g = 9.88605 m s−2 is the gravity of Earth. Since acceleration Actually, gravity on the Earth’s surface varies by around 0.7%, from 9.7639 m s−2 on the Nevado Huascarán mountain in Peru (which is at higher altitude) to 9.8337 m s−2 at the surface of the Arctic Ocean.
is the derivative of velocity, v, it follows that dv dt = g.
According to this equation v just keeps getting larger as a mass continues to fall but we know this isn’t actually true. If you drop a feather, for example, it quickly stops accelerating as air resistance increases until it balances the acceleration due to gravity. The object reaches its terminal velocity.
A skydiver can modify their drag coefficient by sticking out their arms and legs (to increase cd and A), or diving straight down (to decrease cd and A).
It turns out that wind resistance is proportional to v 2 , so that m dv dt = mg −
1 2 ρv 2Acd, where ρ is the density of air, A is the effective area of the falling object, and cd is the drag coefficient (and is dimensionless).
We’ve multiplied the equation by m as that is the usual way of writing it.
Be honest, you know you’ve always wanted to do that.
So let’s suppose you drop a very expensive car, such as a Lamborghini, from a very high place. The car will have a drag coefficient of around 0.35, a mass of around 1,700 kg, and an effective surface area of around 10 m2 . Air has a density of around 1.2 kg/m3 . You don’t throw it, you just drop it, so it starts with zero velocity.

a. Check that the units work out correctly in the differential equation.

b. Use a software package to calculate the velocity of the falling car. What is the terminal velocity (i.e., what is v as t → ∞)?

c. Can you calculate the terminal velocity without solving the differential equation?