Re: "the sound of a large multifunction printer hitting a small white van from a great height"
he did say... "So," I say, "what I need you to do is this: take this printer up to the fourth floor..."
assuming a standard height per floor for buildings of roughly 10ft/floor, which across the pond would equate to the FIFTH floor, with a window height of 3' , but then the target area (van roof) sits about 5' above the ground, so let's just say 48ft (14.63m) . Btw, do not forget to include land elevation above the Thames (assuming that the BOFH is located in London) as a factor of Rho. For this exercise, you may assume that all of London is a flat surface and use an average reading. Points will NOT be deducted from your final answer.
The terminal velocity of a free-falling object is calculated using the formula \(v_t = \sqrt{\frac{2mg}{\rho A C_d}}\), where the key factors are the object's mass (\(m\)), gravitational acceleration (\(g\)), and air density (\(\rho \)), cross-sectional area (\(A\)), and drag coefficient (\(C_{d}\)). It is the steady speed reached when downward gravity equals upward air resistance.
The Formula and Variables
\(v_{t}\): Terminal velocity (m/s)
\(m\): Mass of the object (kg)
\(g\): Acceleration due to gravity (\(9.8\text{ m/s}^2\) on Earth)
\(\rho \) (rho): Density of the fluid/air (\(\approx 1.22\text{ kg/m}^3\) at sea level)
\(A\): Projected cross-sectional area facing the wind (\(\text{m}^{2}\))
\(C_{d}\): Drag coefficient based on shape (dimensionless, e.g., \(\approx 1.0\) for a skydiver belly-down, \(\approx 0.47\) for a sphere)
You now have all relevant information to supply us with an answer. We've even supplied some rudimentary support from https://www.omnicalculator.com/physics/terminal-velocity