A particle of mass `m` is projected with a speed `v_(0)` at an angle of projection `theta` with horizontal. Find the torque of the weight of the particle about the point of projection after time `(v_(0)sintheta)/(g)`

Explanation

In projectile motion,

Time of flight ` T= (2v_(0)sintheta)/(g)`
Horizontal range `R=(v_(0)^(2))/(g) sin2theta`
After time `t=(v_(0)sintheta)/(g)`, the particle is at highest point.
Torque of wieght `mg` about point of projection
`tau_(o)=mgR/2=(mgv_(0)^(2)sin2theta)/(2g)`
`=(1)/(2)mv_(0)^(2)sin2theta`