Year: 2012
Paper: 1
Question Number: 9
Course: LFM Pure and Mechanics
Section: Projectiles
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1488.1
Banger Comparisons: 1
A tall shot-putter projects a small shot from a point $2.5\,$m above the ground, which is horizontal. The speed of projection is $10\,\text{ms}^{- 1}$ and the angle of projection is $\theta$ above the horizontal. Taking the acceleration due to gravity to be $10\,\text{ms}^{-2}$, show that the time, in seconds, that elapses before the shot hits the ground is
\[
\frac1{\sqrt2}\left ( \sqrt{1-c}+ \sqrt{2-c}\right),
\]
where $c = \cos2\theta$.
Find an expression for the range in terms of $c$ and show that it is greatest when $c= \frac15\,$. Show that the extra distance attained
by projecting the shot at this angle rather than at an angle of $45^\circ$ is $5(\sqrt6 -\sqrt2 -1)\,$m.\begin{align*}
&& s &= ut + \frac12 gt^2 \\
\Rightarrow && -2.5 &= 10 \sin \theta \, T - 5 T^2 \\
\Rightarrow && T &= \frac{10\sin \theta \pm \sqrt{100\sin^2 \theta - 4 \cdot 5 \cdot (-2.5)}}{10} \\
&&&= \sin \theta +\sqrt{\sin^2 \theta + \frac12} \\
&&&= \frac1{\sqrt{2}} \left ( \sqrt{2} \sin \theta +\sqrt{2 \sin^2 \theta +1} \right) \\
&&&= \frac1{\sqrt{2}} \left ( \sqrt{2 (1-\cos^2 \theta)} + \sqrt{2-\cos 2\theta} \right) \\
&&&= \frac1{\sqrt{2}} \left ( \sqrt{1-\cos2 \theta} + \sqrt{2-\cos 2\theta} \right) \\
&&&= \frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right)\\
\\
&& s &= 10 \cos \theta T \\
&&&= 10 \sqrt{\frac{\cos 2 \theta +1}{2}}\frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\
&&&= 5 \sqrt{c+1}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\
\\
&& \frac15\frac{\d s}{\d c} &= \frac12(c+1)^{-1/2}((1-c)^{1/2} + (2-c)^{1/2}) - \frac12(c+1)^{1/2}\left ((1-c)^{-1/2}+(2-c)^{-1/2} \right) \\
&&&= \frac{((1-c)(2-c)^{1/2}+(2-c)(1-c)^{1/2})-((c+1)(2-c)^{1/2}+(c+1)(1-c)^{1/2})}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\
&&&= \frac{\sqrt{2-c}\left (1-c-c-1 \right)+\sqrt{1-c}\left(2-c-c-1) \right)}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\
&&&= \frac{\sqrt{1-c}\left(1-2c\right)-2c\sqrt{2-c}}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ \\
\frac{\d s}{\d c} =0: && \sqrt{1-c}\left(1-2c\right)&=2c\sqrt{2-c} \\
\Rightarrow && (1-c)(1-2c)^2&=4c^2(2-c) \\
\Rightarrow && 1-5c+8c^2-4c^3 &= 8c^2-4c^3 \\
\Rightarrow && 0 &= -5c+1 \\
\Rightarrow && c &= \frac15
\end{align*}
When $\theta = 45^{\circ}, c = 0$, so $s_{45^{\circ}} = 5(1+\sqrt{2})$
When $c = \frac15$,
\begin{align*}
s &= 5 \sqrt{\frac15+1}\left ( \sqrt{1-\frac15}+\sqrt{2-\frac15} \right) \\
&= 5 \sqrt{\frac65} \left ( \sqrt{\frac45} + \sqrt{\frac95} \right) \\
&= 2\sqrt{6}+3\sqrt{6} = 5\sqrt{6}
\end{align*}
Therefore the additional distance is $5(\sqrt{6}-\sqrt{2}-1)$
This question was quite popular. The first part of the question involves the application of the formulae for motion under uniform acceleration and this was generally well carried out, although a number of candidates did not justify the choice of the positive square root. A number of candidates also took a longer approach to the calculation, calculating the time to reach the highest point and then the time for the downward journey. Many candidates realised that differentiation of the expression for the range was required, although some decided to differentiate with respect to , rather than , making the task more difficult. The differentiation requires a degree of care to make sure that the signs are correctly managed and many candidates did manage to complete this successfully.