2022TrialShort answer

Q14 (c) (ii)

3 marks

Question

Show that
$x=180\ln\left(\frac{253600}{3600+v^2}\right)$
and find the maximum height the projectile reaches.

Try the question before revealing the answer.

Solution

(ii)

v\frac{dv}{dx}=-10-\frac{v^2}{360}

=-\frac{3600-v^2}{360}

\frac{dv}{dx}=-\frac{3600-v^2}{360v}

\frac{dx}{dv}=-\frac{360v}{3600+v^2}

x=-180\int\frac{2v}{3600+v^2}\,dv

=-180\ln(3600+v^2)+C.

x=0,\quad v=500.

0=-180\ln(3600+500^2)+C

180\ln(253600)=C

x=180\ln\left(\frac{253600}{3600+v^2}\right).

Max. height: $x$ when $v=0$

=180\ln\left(\frac{253600}{3600+0^2}\right)

=765.86\ldots

\therefore 766\,\mathrm{m}\ [\text{nearest m}].

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Syllabus alignment

Mechanics

Resisted motionDerive and solve motion models with resistance on horizontal and vertical paths, including terminal velocity.

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