2022TrialShort answer

Q14 (c) (iii)

2 marks

Question

Show that the time taken for the projectile to reach velocity $v\,\mathrm{m/s}$ is
$t=6\tan^{-1}\left(\frac{25}{3}\right)-6\tan^{-1}\left(\frac{v}{60}\right),$
and find the time taken for the projectile to reach its maximum height.

Try the question before revealing the answer.

Solution

(iii)

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

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

t=\int\frac{-360}{3600+v^2}\,dv

=-360\times\frac{1}{60}\tan^{-1}\left(\frac{v}{60}\right)+C.

t=-6\tan^{-1}\left(\frac{v}{60}\right)+C.

t=0,\quad v=500

0=-6\tan^{-1}\left(\frac{500}{60}\right)+C

=-6\tan^{-1}\left(\frac{25}{3}\right)+C.

6\tan^{-1}\left(\frac{25}{3}\right)=C.

t=6\tan^{-1}\left(\frac{25}{3}\right)-6\tan^{-1}\left(\frac{v}{60}\right).

Time for max height is $t$ when $v=0$ :

=6\tan^{-1}\left(\frac{25}{3}\right)

=8.708\ldots

\text{Time to max. height}=8.7\text{ seconds}

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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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