1. Problem Setup

We have two crumpled paper objects with identical shapes (same aerodynamics) but different masses.

• Object 1 (Heavier): Made of 2 sheets. Mass $M_1 = 2m$.
• Object 2 (Lighter): Made of 1 sheet. Mass $M_2 = m$.

The air resistance force is given by $F_R = C v^x$, where $C$ is a constant depending on the shape (which is identical for both) and air properties.

2. Terminal Velocity Condition

At terminal velocity ($v_t$), the resistive force balances gravity:

$$ Mg = C v_t^x \implies v_t = \left( \frac{Mg}{C} \right)^{1/x} $$

For the heavier object ($v_1$):

$$ v_1 = \left( \frac{2mg}{C} \right)^{1/x} $$

For the lighter object ($v_2$):

$$ v_2 = \left( \frac{mg}{C} \right)^{1/x} $$

3. Relationship given in problem

The problem states that after acquiring terminal velocities, the heavier one moves $\eta = \sqrt{2}$ times farther than the lighter one in the same time interval.

Since velocity is constant, distance $d = v_t \times t$. Thus, the ratio of distances equals the ratio of terminal velocities.

$$ \frac{d_1}{d_2} = \frac{v_1}{v_2} = \eta = \sqrt{2} $$ Substituting the expressions for velocity: $$ \frac{ \left( \frac{2mg}{C} \right)^{1/x} }{ \left( \frac{mg}{C} \right)^{1/x} } = \sqrt{2} $$ $$ (2)^{1/x} = 2^{1/2} $$

4. Conclusion

Equating the exponents:

$$ \frac{1}{x} = \frac{1}{2} $$