Olympiad workbookNSEPLaws of Motion
A particle of mass m is thrown vertically up with velocity u . Air exerts an opposing force of a constant magnitude F . The particle returns back to the point of projection with velocity v after attaining maximum height h , then
Options
- Ah= u ^2 2 ( ~g + F m )
- Bh= v ^2 2 ( ~g - F m )
- Cv=u (g- F m ) (g+ F m )
- Dv=u (g+ F m ) (g- F m )
Correct answer
C. v=u (g- F m ) (g+ F m )
Step-by-step solution
No Solution