Solution

In the laboratory frame, the lifetime of a moving particle is increased due to the relativistic time dilation effect:

$$\tau' = \frac{\tau}{\sqrt{1 - \frac{v^2}{c^2}}}$$

The particle must cover the distance $l$ within the time $\tau'$:

$$l < v \tau'$$

Substituting the expression for $\tau'$:

$$l < \frac{v \tau}{\sqrt{1 - \frac{v^2}{c^2}}}$$

Let's solve this inequality for $v$. First, divide both sides by $\tau$ and square both sides:

$$\frac{v^2}{1 - \frac{v^2}{c^2}} > \left( \frac{l}{\tau} \right)^2$$

Multiply both sides by $1 - \frac{v^2}{c^2}$:

$$v^2 > \left( \frac{l}{\tau} \right)^2 \left( 1 - \frac{v^2}{c^2} \right)$$

Distribute and bring all terms to one side:

$$v^2 + \left( \frac{l}{\tau c} \right)^2 v^2 > \left( \frac{l}{\tau} \right)^2$$

Factor out $v^2$:

$$v^2 \left( 1 + \left( \frac{l}{\tau c} \right)^2 \right) > \left( \frac{l}{\tau} \right)^2$$

Now solve for $v$:

$$v^2 > \frac{\left( \frac{l}{\tau} \right)^2}{1 + \left( \frac{l}{\tau c} \right)^2}$$

$$v > \frac{c}{\sqrt{1 + \left( \frac{\tau c}{l} \right)^2}}$$