A ball of mass m is thrown upward with a velocity `upsilon`. If air exerts an average resisting force F, the velocity with which the ball returns to the thrower is

Options

A. `v sqrt((mg)/(mg+F))`
B. `v sqrt((F)/( mg +F))`
C. `v sqrt((mg -F)/( mg +F))`
D. `v sqrt((mg +F)/( mg ))`

Answer

C

Explanation

For upward motion
Net force `=mg +F`
` therefore ` Retardation `(a) =(mg +F)/( m)`
Distance,s `= (v^2 )/(2a) = (v^2 m)/( 2(mg +F))`
For downward motion , net force `= mg - F `
` therefore ` Acceleration `(a.) = ( Mg-F)/( m)`
distance ,s .` (v^2.)/( 2 a .) = (v^2. m)/(2 ( mg -F))`
As ` s=s. therefore (v^2 m)/( 2 (mg +F))=(v.^2 m)/(2 (mg -F))`
` or v.=v sqrt((mg -F)/( mg +F))`