A 75 Kg Skydiver Can Be Modeled as a Rectangular
Skydiver elevate problem
- Thread starter vau
- Start engagement
A 75.0 kg skydiver can exist modeled as a rectangular "box" with dimensions 24.0 cm\times 44.0 cm\times 184 cm.
What is his terminal speed if he falls feet first?
What is his terminal speed if he falls feet first?
I know that drag = mg when terminal speed is reached. and then elevate is 75*9.8
and at that place's a formula drag = KV^2 where K is elevate coeff and V is velocity, but I don't know One thousand.
a piddling aid?
Answers and Replies
Okay, so you summed the forces when they were in equilibrium and found the forcefulness of elevate. You don't know K, does that mean yous are supposed to derive it? If so, my hint is that you were given two different quantities in your trouble, and you lot already used one... so K must have something to exercise with the other.
[tex]K = \frac{1}{ii} \rho Southward C_D[/tex]
Where
[tex]\rho[/tex] is air density.
Due south is the cross section expanse. In your case 24cm x 44cm.
[tex]C_D[/tex] is the elevate coefficient, unremarkably determined experimentally. If you don´t accept it´south value I don´t see how you could solve the problem.
Where
[tex]\rho[/tex] is air density.
Due south is the cross section expanse. In your case 24cm x 44cm.
[tex]C_D[/tex] is the elevate coefficient, unremarkably determined experimentally. If you don´t accept it´south value I don´t see how you could solve the problem.
Related Threads on Skydiver elevate problem
- Concluding Postal service
- Last Post
- Concluding Post
- Last Post
- Final Postal service
- Last Post
- Last Post
- Final Postal service
- Terminal Mail service
- Final Post
Source: https://www.physicsforums.com/threads/skydiver-drag-problem.133595/
0 Response to "A 75 Kg Skydiver Can Be Modeled as a Rectangular"
Post a Comment