A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be (Take radius of earth $$= 6400 \text{ km}$$ and $$g = 10 \text{ m s}^{-2}$$)

A body is projected vertically upwards with velocity $$v = \dfrac{v_e}{3}$$ (one-third of escape velocity), and we need to find the maximum height attained. The escape velocity from Earth's surface is $$v_e = \sqrt{2gR}$$ where $$R = 6400 \text{ km}$$ and $$g = 10 \text{ m s}^{-2}$$.

At the surface, the total energy equals kinetic plus potential energy: $$\dfrac{1}{2}mv^2 - \dfrac{GMm}{R} = -\dfrac{GMm}{R+h}$$.

Since $$v = \dfrac{v_e}{3} = \dfrac{\sqrt{2gR}}{3}$$, substituting into the energy equation yields $$\dfrac{1}{2}m \cdot \dfrac{2gR}{9} = \dfrac{GMm}{R} - \dfrac{GMm}{R+h}$$.

Using $$GM = gR^2$$, this gives $$\dfrac{mgR}{9} = mgR\left(1 - \dfrac{R}{R+h}\right) = mgR \cdot \dfrac{h}{R+h}$$. From this, $$\dfrac{1}{9} = \dfrac{h}{R+h}$$ so that $$R + h = 9h$$ and $$R = 8h$$, which leads to $$h = \dfrac{R}{8} = \dfrac{6400}{8} = 800 \text{ km}$$.

The correct answer is Option A: $$800 \text{ km}$$.