A stone with weight W is thrown vertically upward into the air with initial velocity `v_(0)`. If a constant forcef due to air drag acts on the stone throughout the flight & if the maximum height attain by stone is h and velocity when it strikes to the ground is v. Which one is correct?

Options

A. `v_0((w-f)/(w+f))^(1//2)`
B. `v_0((w+f)/(w-f))^(1//2)`
C. `v_0((w-f)/(w))^(1//2)`
D. `v_0((w+f)/(f))^(1//2)`

Answer

A

Explanation

During rise of stone , retardation = (g+a)
where `g = w/m and a = f/m `
During fall of stone , acceleration = (g -a)
Let the height reached by stone be h .
For upward journey up to the highest point,
`0 =v_(0)^(2) -2 (g +a)h " or " 2h (g+a) =v_0^2`
For downward journey from this height ,
`v^(2) = 0 =2 (g - a) h " or " 2h (g -a ) = v^(2)`
`:. (v^2)/(v_0^2) = (2h(g-a)/(2h(g+a)) = (g-a)/(g+a) = (m(g-a))/(m(g+a))=(mg - ma )/(mg +ma)`
or `v_0^2/v^2 =(w-f)/(w+f) " or " v = v_0 ((w-f)/(w+f))^(1//2)`