STEP III 2007 question 10 solution

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STEP III 2007 question 10 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 10 solution

Draw the obligatory diagram, and we see that in the (x,y) coord frame, acceleration due to gravity is $(-g \sin \phi, -g \cos \phi)$ .

Thus the equations for x and y are:

$x = Vt \cos \theta - \frac{1}{2}gt^2 \sin \phi, \quad y = Vt \sin \theta - \frac{1}{2}gt^2 \cos \phi$

When the particle hits the ground, y=0 (and $t\neq 0$ ), so $V \sin \theta = \frac{1}{2}gt \cos \phi$ . [B](1)[/B]

The rebound only affects the y component of the velocity, so if it is to retrace its path, we must have $\dot{x}= 0$ and so $V \cos \theta = gt \sin \phi$ . [b](2)[/b]

Divide (2) by (1) to get $\cot \theta = 2 \tan \phi$ and so $2 \tan \theta \tan \phi = 1$ [b](3)[/b].

Rewrite (2) to get $t=\frac{V \cos \theta}{g \sin \phi}$ at time of landing. Substitute into the equation for x to get the maximum range:

$\displaystyle R = \frac{V^2 \cos^2 \theta}{g \sin \phi} - \frac{g}{2} \frac{V^2 \cos^2 \theta}{g^2 \sin^2 \phi} \sin \phi = \frac{V^2 \cos^2 \theta}{2g \sin \phi}$ as required.

So then $\displaystyle \frac{2V^2}{gR} = \frac{2V^2}{g} \frac{2g \sin \phi}{V^2 \cos^2 \theta}$

$= 4 \sin \phi \sec^2 \theta$

$= \sin \phi (4 + 4 \tan^2 \theta)$ .

$=\sin\phi(4+\cot^2 \phi)$ (Using (3) to tell us $2\tan \theta = \cot \phi$ ).

$=\sin \phi (3+\textrm{cosec}^2 \phi) = 3 \sin \phi + \textrm{cosec} \phi$ as required.

Now $\displaystyle 3 \sin \phi + \textrm{cosec} \phi = \sqrt{3}\left(\sqrt{3}\sin \phi + \frac{1}{\sqrt{3}\sin \phi}\right)$

$\displaystyle = \sqrt{3}\left(2+\left(\sqrt{\sqrt{3}\sin \phi} - \sqrt{\frac{1}{\sqrt{3}\sin \phi}\right)^2\right)$

$\geq 2\sqrt{3}$ , with equality when $\sin \phi = \frac{1}{\sqrt{3}}$ .

Thus $\frac{2V^2}{gR} \geq 2\sqrt{3}$ with equality when $\sin \phi = \frac{1}{\sqrt{3}}$ .

Then $R \leq \frac{V^2}{g\sqrt{3}}$ with equality when $\sin \phi = \frac{1}{\sqrt{3}}$ .

Solution by DFranklin.

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