A tennis player serves from height \(H\) above horizontal ground, hitting the ball downwards with speed \(v\) at an angle \(\alpha\) below the horizontal. The ball just clears the net of height \(h\) at horizontal distance \(a\) from the server and hits the ground a further horizontal distance \(b\) beyond the net. Show that $$ v^2 = \frac{ g(a+b)^2(1+\tan^2\alpha)}{ 2[H-(a+b)\tan\alpha]} $$ and $$ \tan\alpha = \frac{2a+b }{ a(a+b)}H - \frac{a+b }{ ab}h \,. $$ By considering the signs of \(v^2\) and \(\tan\alpha\), find upper and lower bounds on \(H\) for such a serve to be possible.
The game of Cambridge Whispers starts with the first participant Albert flipping an un-biased coin and whispering to his neighbour Bertha whether it fell `heads' or `tails'. Bertha then whispers this information to her neighbour, and so on. The game ends when the final player Zebedee whispers to Albert and the game is won, by all players, if what Albert hears is correct. The acoustics are such that the listeners have, independently at each stage, only a probability of 2/3 of hearing correctly what is said. Find the probability that the game is won when there are just three players. By considering the binomial expansion of \((a+b)^n+(a-b)^n\), or otherwise, find a concise expression for the probability \(P\) that the game is won when is it played by \(n\) players each having a probability \(p\) of hearing correctly. % Show in particular that, if \(n\) is even, %\(P(n,1/10) = P(n,9/10)\).% How do you explain this apparent anomaly? To avoid the trauma of a lost game, the rules are now modified to require Albert to whisper to Bertha what he hears from Zebedee, and so keep the game going, if what he hears from Zebedee is not correct. Find the expected total number of times that Albert whispers to Bertha before the modified game ends. \noindent [You may use without proof the fact that \(\sum_1^\infty kx^{k-1}=(1-x)^{-2}\) for \(\vert x\vert<1\).]
\item A needle of length two cm is dropped at random onto a large piece of paper ruled with parallel lines two cm apart.
Traffic enters a tunnel which is 9600 metres long, and in which overtaking is impossible. The number of vehicles which enter in any given time is governed by the Poisson distribution with mean 6 cars per minute. All vehicles travel at a constant speed until forced to slow down on catching up with a slower vehicle ahead. I enter the tunnel travelling at 30 m\(\,\)s\(^{-1}\) and all the other traffic is travelling at 32 m\(\,\)s\(^{-1}\). What is the expected number of vehicles in the queue behind me when I leave the tunnel? Assuming again that I travel at 30 m\(\,\)s\(^{-1}\), but that all the other vehicles are independently equally likely to be travelling at 30 m\(\,\)s\(^{-1}\) or 32 m\(\,\)s\(^{-1}\), find the probability that exactly two vehicles enter the tunnel within 20 seconds of my doing so and catch me up before I leave it. Find also the probability that there are exactly two vehicles queuing behind me when I leave the tunnel. \noindent [Ignore the lengths of the vehicles.]