Given: A body is projected vertically upward from the surface of Earth and must reach a height of $10R$ above the surface.

Interpret the final distance from Earth's center:

• Initial distance from Earth's center: $R$
• Final height above surface: $10R$
• So final distance from center is $r = R + 10R = 11R$

Use conservation of mechanical energy: Let the required initial speed be $v_i$.

Initial total energy at Earth's surface: $E_i = \frac{1}{2}mv_i^2 - \frac{GMm}{R}$

At the maximum height $11R$, the body just reaches there, so final speed is $0$: $E_f = -\frac{GMm}{11R}$

By conservation of energy, $\frac{1}{2}mv_i^2 - \frac{GMm}{R} = -\frac{GMm}{11R}$

**Solve for $v_i$: ** $\frac{1}{2}mv_i^2 = \frac{GMm}{R} - \frac{GMm}{11R}$ $\frac{1}{2}mv_i^2 = GMm\left(\frac{1}{R} - \frac{1}{11R}\right)$ $\frac{1}{2}mv_i^2 = GMm\left(\frac{10}{11R}\right)$

Therefore, $v_i^2 = 2GM\cdot \frac{10}{11R} = \frac{20GM}{11R}$

Use escape velocity relation: Escape velocity from Earth's surface is $v_e = \sqrt{\frac{2GM}{R}}$ so $v_e^2 = \frac{2GM}{R}$

Hence, $v_i^2 = \frac{10}{11}v_e^2$ $v_i = \sqrt{\frac{10}{11}}\,v_e$

Compare with the given form: $v_i = \sqrt{\frac{x}{y}}\,v_e$ Thus, $x=10, \quad y=11$

Required value: $\boxed{x=10}$