Question

A particle of mass $m$ is projected at time $t=0$ with a velocity $u$ making an angle ${45}^0$ with the horizontal. The magnitude of the torque due to weight of the particle is $\left(\frac{m{u}^2}{x}\right)$, when it is at its maximum height, about a point where the particle was at time = $\frac{u}{2\sqrt{2}g}$ on the trajectory. Find the value of $x$.

Answer 1

time to reach highest point $usin\left(45\right)=gt\to t=\frac{u}{g\sqrt{2}}$

horizontal displacement when it is at highest point $D_1=ucos45\times t=\frac{u^2}{2g}$

when $t_1=\frac{u}{2\sqrt{2}g}$

$D_2=ucos25\times t_2=\frac{u^2}{4g}$

$Torque=mg(D_1-D_2)=\frac{m{u}^2}{4}$

$\therefore x=4$