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A body of mass m is thrown upwards at an...

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

A

`sqrt((vcostheta)^2+(vsintheta)^2)`

B

`sqrt((vcostheta-vsontheta^2)-g t)`

C

`sqrt(v^2+g^2t^2-(2vsintheta)g t)`

D

`sqrt(v^2+g^2t^2-(2vcostheta)g t)`

Text Solution

Verified by Experts

The correct Answer is:
C
Instantaneous velocity of rising mass after t sec will be
`v_t=sqrt(v_x^2+v_y^2)`
where `v_x=vcostheta=`horizontal component of velocity
`v_y=vsintheta-g t=` vertical component of velocity
`v_t=sqrt((vcostheta)^2+(vsintheta-g t)^2)`
`v_t=sqrt(v^2+g^2t^2-2vsinthetag t)`
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