Understanding the Problem
To determine the minimum number of plants required to stop a bullet, we first analyze the situation where the bullet loses 1/50 of its velocity upon passing through each plant.

Initial Velocity and Loss Calculation
- Let the initial velocity of the bullet be \( v_0 \).
- After passing through the first plant, the bullet’s velocity reduces to:
\[
v_1 = v_0 - \frac{1}{50}v_0 = \frac{49}{50}v_0
\]
- After passing through the second plant, the velocity becomes:
\[
v_2 = v_1 - \frac{1}{50}v_1 = \frac{49}{50}v_1 = \left(\frac{49}{50}\right)^2 v_0
\]
- Continuing this pattern, after passing through \( n \) plants, the velocity will be:
\[
v_n = \left(\frac{49}{50}\right)^n v_0
\]

Condition for Stopping the Bullet
- The bullet stops when its velocity becomes zero. Mathematically, this can be expressed as:
\[
v_n \leq 0
\]
- However, practically, we consider it to be negligible (almost zero). Thus, we focus on the condition:
\[
\left(\frac{49}{50}\right)^n v_0 < />
\]
where \( \epsilon \) is a small value close to zero.

Solving for \( n \)
- Taking natural logarithms on both sides:
\[
n \ln\left(\frac{49}{50}\right) < />
\]
- Rearranging gives:
\[
n > \frac{\ln\left(\frac{\epsilon}{v_0}\right)}{\ln\left(\frac{49}{50}\right)}
\]
- Since \( \ln\left(\frac{49}{50}\right) \) is negative, we take the absolute value.

Conclusion
To find the minimum number of plants \( n \) required to stop the bullet, compute the above inequality using values for \( \epsilon \) and \( v_0 \). Hence, the required number of plants can be determined based on initial conditions.