Solution to Question 30
1. Differential Equation for Horizontal Motion
The only force acting horizontally is the air drag $F = -kv$.
$$m \frac{dv_x}{dt} = -k v_x$$ Integrate to find velocity as a function of time: $$\int_{v_0}^{v} \frac{dv}{v} = -\frac{k}{m} \int_{0}^{t} dt \implies v(t) = v_0 e^{-\frac{kt}{m}}$$2. Determine Horizontal Range
The range $R$ is the distance covered in time $\tau$.
$$x = \int_{0}^{\tau} v(t) dt = \int_{0}^{\tau} v_0 e^{-\frac{kt}{m}} dt$$ $$R = v_0 \left[ -\frac{m}{k} e^{-\frac{kt}{m}} \right]_{0}^{\tau} = \frac{m v_0}{k} \left( 1 – e^{-\frac{k\tau}{m}} \right)$$3. Substitute Initial Energy
Given kinetic energy $K = \frac{1}{2}mv_0^2 \implies v_0 = \sqrt{\frac{2K}{m}}$. Substituting this into the range equation:
$$R = \frac{m}{k} \sqrt{\frac{2K}{m}} \left( 1 – e^{-\frac{k\tau}{m}} \right)$$ $$R = \frac{\sqrt{2Km}}{k} \left( 1 – \exp\left(-\frac{k\tau}{m}\right) \right)$$