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The problem you've uploaded appears to be about the motion of a particle fired vertically upward, where the particle experiences air resistance proportional to the square of its velocity, alongside the gravitational force. The resistance force is given by:

\[
|F| = 0.098 m v^2
\]

You are asked to show that the height \( y \) as a function of velocity \( v \) is:

\[
y = \frac{250}{49} \log_e \left( \frac{100 - u^2}{100 - v^2} \right)
\]

with the assumption that \( g = 9.8 \, \text{m/s}^2 \).

### Approach to solve:

1. **Forces on the particle:**
   - The total force on the particle includes gravitational force \( mg \) and air resistance proportional to the square of the velocity.
   
2. **Net Force and Newton's Second Law:**
   Use Newton's second law to form the differential equation of motion for the particle:
   
   \[
   m \frac{dv}{dt} = -mg - 0.098 m v^2
   \]
   
   Simplifying:
   \[
   \frac{dv}{dt} = -g \left(1 + 0.01 v^2 \right)
   \]

3. **Separating Variables:**
   This can then be rearranged to separate the variables \( v \) and \( t \), and we can proceed to integrate to find the relationship between velocity and height \( y \). The integration will involve logarithmic terms as hinted by the result to prove.

Would you like to continue with a full step-by-step derivation of this process? 

Here are some related questions:
1. How do you solve differential equations involving air resistance proportional to \(v^2\)?
2. How do logarithmic terms arise when solving integrals of this form?
3. What role does Newton's second law play in problems involving variable forces?
4. How can the gravitational force and air resistance affect the maximum height a particle reaches?
5. What assumptions are made when dealing with air resistance proportional to velocity squared?

**Tip:** When handling forces involving air resistance, the equation becomes nonlinear, often leading to logarithmic solutions after integration.

A particle is fired vertically upwards from the ground with an initial velocity u. It experiences a force from air resistance proportional to the square of its velocity, |F| = 0.098m v^2, as well as the gravitational force. Show that: y = (250/49) * log_e((100 - u^2) / (100 - v^2)). You may assume g = 9.8 ms^(-2).

Solve a vertical motion problem involving a particle experiencing both air resistance proportional to velocity squared and gravitational force. Learn how to derive the height equation y = (250/49) * log_e((100 - u^2) / (100 - v^2)) using Newton's Second Law and separation of variables.

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