A particle is projected with velocity $$v_0$$ along $$x$$-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. $$ma = -\alpha x^2$$. The distance at which the particle stops:

The particle starts with velocity $$v_0$$ along the x-axis and experiences a damping force given by $$ma = -\alpha x^2$$. Writing acceleration as $$a = v\frac{dv}{dx}$$, we get $$mv\,dv = -\alpha x^2\,dx$$.

Integrating both sides, $$m\int_{v_0}^{0} v\,dv = -\alpha \int_{0}^{x_0} x^2\,dx$$, where $$x_0$$ is the distance at which the particle stops ($$v = 0$$).

Evaluating the integrals, $$m\left[\frac{v^2}{2}\right]_{v_0}^{0} = -\alpha\left[\frac{x^3}{3}\right]_{0}^{x_0}$$, which gives $$-\frac{mv_0^2}{2} = -\frac{\alpha x_0^3}{3}$$.

Simplifying, $$\frac{mv_0^2}{2} = \frac{\alpha x_0^3}{3}$$, so $$x_0^3 = \frac{3mv_0^2}{2\alpha}$$.

Therefore, the distance at which the particle stops is $$x_0 = \left(\frac{3mv_0^2}{2\alpha}\right)^{\frac{1}{3}}$$.

The correct answer is $$\left(\frac{3mv_0^2}{2\alpha}\right)^{\frac{1}{3}}$$.