A ball is thrown upward with an initial velocity `V_0` from the surface of the earth.The motion of the ball is affected by a drag force equal to `mgammav^2` (where m is mass of the ball, v is its instantaeous velocity and `gamma` is a constant).Time taken by the ball to rise to its zenith is :

Options

A. `1/(sqrt(gammag))tan^(-1)(sqrt((gamma)/g)v_(0))`
B. `1/(sqrt(gammag))In(1+sqrt((gamma)/g)v_(0))`
C. `1/(sqrt(gammag))sin^(-1)(sqrt(gamma)/gv_(0))`
D. `1/(sqrt(2)gammag)tan^(-1)(sqrt((2gamma)/g)v_(0))`

Answer

A

Explanation

`a = -(g +gammav^(2))`
`(dv)/(dt) =-(g+gammav^(2))rArrint_(v_(0))^(0) (dv)/(g+gammav^(2))=-int_(0)^(t) dt`
` 1/gamma int_(v_(0))^(0)(dv)/((g/gamma+v^(2)))=-int_(0)^(1)dt rArr1/gamma 1/(sqrt(g/gamma))[tan^(-1)(v/(sqrt(g/gamma)))]_(v_(0))^(0)=-t`
`1/(sqrt(ggamma))tan^(-1)((sqrt(gamma))/(sqrt(g))v_(0))=t`