Question

A particle having mass $m$ is projected with a velocity $v_0$ from a point $P$ on a horizontal ground making an angle $\theta$ with horizontal. Find out the torque about the point of projection acting on the particle when it is at its maximum height?

• A. $\frac{m{v}_0^{2}sin2\theta }{4}$
• B. $\frac{m{v}_0^{2}sin2\theta }{2}$
• C. $\frac{m{v}_0^{2}si{n}^2\theta }{2}$
• D. $\frac{2m{v}_0^{2}sin2\theta }{2}$

Answer 1

The correct option is B

$\frac{m{v}_0^{2}sin2\theta }{2}$

$\tau$ = $Frsin\theta$

As $rsin\theta$ is nothing but the horizontal component of $r$ which is half of the Range, R.
$\tau =Frsin\theta =mg\frac{R}{2}=\frac{v_0^{2}sin2\theta }{2g}mg$
$\tau =\frac{m{v}_0^{2}sin2\theta }{2}$