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 beA. `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)`

A body of mass m is thrown upwards at an angle `theta` with the horizo

1.	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 beA. `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» Correct Answer - C 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)`