 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/15/a/solution

First solve for the velocity components. The integrating factor

e^{γ t}

gives

\overset{x}{˙} = V_{0 x} e^{- γ t}, \overset{z}{˙} = (V_{0 z} + \frac{g}{γ}) e^{- γ t} - \frac{g}{γ} .

Integrating once more and imposing the initial position gives

x (t) = \frac{V _{0 x}}{γ} (1 - e^{- γ t}), z (t) = \frac{V _{0 z} + g / γ}{γ} (1 - e^{- γ t}) - \frac{g}{γ} t .

These formulas describe projectile motion with linear drag. Differentiation checks both second-order differential equations and both initial velocities. At large positive time, the horizontal velocity tends to zero and the vertical velocity tends to the terminal value

- g / γ

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