If a body of mass 1 kg falls on the earth from infinity, it attains...

By conservation of mechanical energy, the total energy of the body at infinity is zero.

When the body reaches the surface of the earth, its potential energy is $-\dfrac{GMm}{R}$ and kinetic energy is $k = \dfrac{1}{2}mv^2$.

$0 = k - \dfrac{GMm}{R}$

$k = \dfrac{GMm}{R} = mgR$

Substituting $m = 1$ kg, $g = 9.8$ m/s$^2$, and $R = 6400 \times 10^3$ m:

$k = 1 \times 9.8 \times 6400 \times 10^3 = 6.272 \times 10^7$ J

The velocity $v$ is given by $\dfrac{1}{2}mv^2 = k$.

$v = \sqrt{2gR} = \sqrt{2 \times 9.8 \times 6400 \times 10^3}$

$v = \sqrt{125.44 \times 10^6} = 11.2 \times 10^3$ m/s $= 11.2$ km/s

Answer: $11.2$ km/s; $6.27 \times 10^7$ J