Let's break down each part of the question step-by-step: --- ### (i) **Force Diagram** When a particle of mass \( m \) is thrown vertically upward with velocity \( v \), the forces acting on it are: - **Gravity (\( mg \)):** Acts downward. - **Air resistance (\( kv^2 \)):** Acts downward (opposite to the direction of motion), where \( k \) is a constant and \( v \) is the upward velocity. **Force Diagram:** ``` ↑ │ │ │ Air resistance (kv^2, downward) │ ↓ ●———●——— Particle (mass m, upward velocity v) │ │ Gravity (mg, downward) ↓ ``` --- ### (ii) **Differential Equation of Motion** Let the upward direction be positive. - **Weight**: \( -mg \) (downward) - **Air resistance**: \( -kv^2 \) (downward, since it opposes the motion) **Net force on the particle:** \[ F = -mg - kv^2 \] By Newton’s Second Law (\( F = ma = m \frac{dv}{dt} \)): \[ m \frac{dv}{dt} = -mg - kv^2 \] \[ \boxed{\frac{dv}{dt} = -g - \frac{k}{m}v^2} \] --- ### (iii) **Rest of the question (not visible)** If you want to continue further (e.g., find the time to reach maximum height), let me know! --- ## **Summary** ### **(i) Force diagram:** - Upward: Initial velocity \( v \) - Downward: Gravity (\( mg \)), Air resistance (\( kv^2 \)) ### **(ii) Differential equation:** \[ \boxed{\frac{dv}{dt} = -g - \frac{k}{m}v^2} \] If you want to solve for time to reach maximum height, let me know!