A particle of mass m is launched with a speed u at an angle θ...

At the highest point of the trajectory, the horizontal distance of the particle from the point of projection is half of the horizontal range, i.e., $x = \dfrac{R}{2}$.

The horizontal range is given by $R = \dfrac{u^2 \sin 2\theta}{g} = \dfrac{2 u^2 \sin\theta \cos\theta}{g}$.

Thus, $x = \dfrac{u^2 \sin\theta \cos\theta}{g}$.

The force acting on the particle is its weight, $F = mg$, which acts vertically downwards.

The magnitude of the torque of the weight about the point of projection is the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force.

$\tau = F \times x = mg \times$\left(\dfrac{u^2 \sin\theta \cos\theta}{g}\right)

$\tau = m u^2 \sin\theta \cos\theta$