Physics I question Skydiver falling

[bpatel4116]

Homework Statement

A 79 kg skydiver can be modeled as a rectangular "box" with dimensions 25 cm x 39 cm x 1.8 m.

What is his terminal speed if he falls feet first?

Homework Equations

The Attempt at a Solution

So I am not sure what I am doing wrong, here is what i did.

D= 1/4 av2 = .25(.25 x .39)= .020

skydiver's weight i took from kg to N which ended up being 774.2

then i did 774.2/.020 and did square root, and i ended up with 196.7, apparently it is saying this is not the right answer, did i do this correctly!? help! :(

 

[djeitnstine]

What formula are you using? isn't the formula for drag [tex]\frac{1}{4}\rho ADv^2[/tex] maybe [tex]\frac{1}{2}\rho ADv^2[/tex] (Its late and I feel lazy to find out which)

where [tex]\rho[/tex] is the density of air, A is the area of the object and D is the coefficient of drag

 

[nlightner]

As a scientist, it's important to always double check your calculations and make sure you are using the correct formulas for the given situation. In this case, the formula for terminal speed is v = √(2mg/ρAC), where m is the mass of the skydiver, g is the acceleration due to gravity, ρ is the density of air, A is the cross-sectional area of the skydiver, and C is the drag coefficient.

Using the given dimensions of the skydiver, we can calculate the cross-sectional area to be 0.009 m². The density of air at sea level is approximately 1.2 kg/m³. Plugging in these values along with the given mass of 79 kg and the acceleration due to gravity of 9.8 m/s², we get a terminal speed of approximately 55 m/s.

It's also important to note that this calculation assumes the skydiver is in a stable, belly-to-earth position. If the skydiver is in a different position, the drag coefficient and cross-sectional area will be different, resulting in a different terminal speed.

In conclusion, it's important to carefully consider all the variables and use the correct formulas when solving physics problems.