 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/b/solution

V_{0 x} \neq = 0

, the first formula in (a) gives

e^{- γ t} = 1 - γ x / V_{0 x}

and

t = - \frac{1}{γ} ln (1 - \frac{γ x}{V _{0 x}}) .

Substitute this and

1 - e^{- γ t} = γ x / V_{0 x}

into

z (t)

z (x) = (\frac{V _{0 z}}{V _{0 x}} + \frac{g}{γ V _{0 x}}) x + \frac{g}{γ ^{2}} ln (1 - \frac{γ x}{V _{0 x}}) .

For finite physical times

t \geq 0

, the domain is

0 \leq γ x / V_{0 x} < 1

, valid for either sign of

V_{0 x}

. If

V_{0 x} = 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/ γ

terms and gives

z \to (V_{0 z} / V_{0 x}) x - g x^{2} / (2 V_{0 x}^{2})

γ \to 0

 New to topics? Read the docs here!