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(D) The torque $\vec{	au}$ about the point of projection is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.Here,the force $\vec{F}$ is the weight $m\

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$\frac{1}{2} m u^2$

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(D) The torque $\vec{	au}$ about the point of projection is given by $\vec{	au} = \vec{r} 	imes \vec{F}$.Here,the force $\vec{F}$ is the weight $m\vec{g}$ acting downwards.At maximum height,the horizontal distance (range) from the point of projection is $x = R/2$,where $R$ is the horizontal range.The formula for horizontal range is $R = \frac{u^2 \sin(2	heta)}{g}$.Given $	heta = 45^o$,$R = \frac{u^2 \sin(90^o)}{g} = \frac{u^2}{g}$.At maximum height,the horizontal position is $x = \frac{R}{2} = \frac{u^2}{2g}$.The vertical force is $F = mg$ acting downwards.The torque magnitude is $	au = F \cdot x = (mg) \cdot \left( \frac{u^2}{2g} ight) = \frac{1}{2} m u^2$.

$A$ particle of mass $m$ is projected with a velocity $u$ making an angle $45^o$ with the horizontal. The magnitude of the torque due to the weight of the projectile,when the particle is at its maximum height,about the point of projection is:

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