Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 15

First solve for the velocity components. The integrating factor $e^{\gamma t}$ givesIntegrating once more and imposing the initial position givesThese 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 $-g/\gamma$.

If $V_{0x}\ne 0$, the first formula in (a) gives $e^{-\gamma t}=1-\gamma x/V_{0x}$ andSubstitute this and $1-e^{-\gamma t}=\gamma x/V_{0x}$ into $z(t)$:For finite physical times $t\ge 0$, the domain is $0\le \gamma x/V_{0x}<1$, valid for either sign of $V_{0x}$. If $V_{0x}=0$, the trajectory is vertical, $x(t)=0$, and in general no single-valued graph $z(x)$ describes it. As a consistency check, the logarithmic expansion at fixed $x$ cancels the apparent $1/\gamma$ terms and gives $z\to (V_{0z}/V_{0x})x-g{x}^2/(2V_{0x}^2)$ as $\gamma \to 0$, the drag-free parabola.