1. Horizontal projectile formula

For a body thrown horizontally with speed $u$ from height $h$:

• Time to reach ground is $t=\sqrt{\frac{2h}{g}}$
• Horizontal distance is $R=ut=u\sqrt{\frac{2h}{g}}$

2. First body

Given:

• horizontal speed $=v$
• height $=H$
• range $=100\,\text{m}$

So, $100=v\sqrt{\frac{2H}{g}} \qquad (1)$

3. Second body

Given:

• mass $=2M$ (mass does not affect projectile motion)
• horizontal speed $=\frac{v}{2}$
• height $=4H$

Time of flight: $t'=\sqrt{\frac{2(4H)}{g}}=\sqrt{\frac{8H}{g}}=2\sqrt{\frac{2H}{g}}$

Horizontal distance: $R'=\frac{v}{2}\cdot 2\sqrt{\frac{2H}{g}}=v\sqrt{\frac{2H}{g}}$

Using equation (1), $R'=100\,\text{m}$

4. Final answer

The second body touches the ground at a distance of $\boxed{100\text{ m}}$