A body of mass m is thrown upwards at an angle `theta` with the horizontal with velocity v. While rising up the velocity of the mass after t second will be

Options

A. `sqrt((v cos theta^(2)) + (v sin theta)^(2))`
B. `sqrt((v cos theta - v sin theta)^(2) - g t)`
C. `sqrt(v^(2) + g^(2)t^(2) - (2b sin theta) g t)`
D. `sqrt(v^(2) + g^(2)t^(2) -(2v cos theta) g t)`

Answer

C

Explanation

Instantaneous velocity of rising mass after ts will be
`v_(t) = sqrt(v_(x)^(2)+v_(y)^(2))`
where, `v_(x) = v cos theta =` Horizontal component of velocity
`v_(y) = v sin theta - g t = ` Vertical component of velocity
`rArr v_(t) = sqrt((v cos theta)^(2)+(v sin theta - g t)^(2))`
`rArr v_(t) = sqrt(v^(2)+g^(2)t^(2)-(2v sin theta)g t)`