Calculating Terminal Velocity: Physics Help for a Mock Trial Project

AI Thread Summary
To calculate the terminal velocity of a person falling from 100 meters, it's essential to consider air resistance, which significantly affects the outcome. The initial estimation without air resistance is around 100 miles per hour or 44 meters per second. Air drag is proportional to the square of the velocity and depends on the object's shape, with a drag coefficient influencing the calculations. A formula for velocity as a function of height is provided, incorporating mass and the air resistance constant. Understanding these principles will help accurately determine the impact velocity for the mock trial project.
honeydukes
Messages
2
Reaction score
0
Hi. I could use some help from someone who is pretty good at physics.

Me and a few others are doing a project (that has nothing to do with physics) and we need to know the velocity of a person at the time of impact who fell from 100 meters. We figured out that it would be about 100miles per hour/44meters per second without air resistance, but we need to know what it would be WITH air resistance. This person is about 5'10'' and weighs about 150lbs/68.04kg. It's for a sort of mock trial. If anyone could help, we'de really appreciate it!
 
Physics news on Phys.org
At that short of a distance, I doubt that air drag has much of any effect. Basically, air drag is proportional to the square of velocity. The proportionality constant (called drag coefficient) is dependent mainly on the shape of the object. Example - an arrow has minimal air drag compared to a sphere of the same volume.

Anyway, if you want to dig into it, Google searches with terms like 'terminal velocity', 'air drag', 'drag coefficient', will probably get you everything you wanted to know.
 
Assuming you jump off from rest, and that air resistance is modeled as proportional to the square of the velocity, it is relatively easy to derive the velocity dependence on h, the distance fallen from the jump off point:
v(h)=\frac{mg}{k}\sqrt{1-e^{-\frac{2kh}{m}}}
where m is the object's mass, k the air resistance proportionality constant.
Note that in the limit \frac{kh}{m}\to{0}[/tex], we get v(h)=\sqrt{2gh}[/tex] as we of course should have.<br /> (This include the special cases of zero air resistance or infinite mass of the falling object)
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top