A projectile is fired from the origin O with initial velocity V m s^-1 at an angle θ to the horizontal - HSC - SSCE Mathematics Extension 1 - Question 14 - 2015 - Paper 1

Given that $\theta = \frac{\pi}{3}$, we can calculate the vertical position:

1. Using the vertical motion equation, we find:

$y = V \sin \theta \cdot t - \frac{1}{2}gt^2 = V \sin \left(\frac{\pi}{3}\right) \cdot \frac{2V}{\sqrt{3g}} - \frac{1}{2}g \left(\frac{2V}{\sqrt{3g}}\right)^2.$

2. Solving this gives us:

$y = \frac{\sqrt{3}}{2}V \cdot \frac{2V}{\sqrt{3g}} - \frac{2V^2}{3} = \frac{V^2}{g} - \frac{2V^2}{3} = \frac{V^2}{3g}.$

3. The horizontal displacement is:

$x = V \cos \left(\frac{\pi}{3}\right) \cdot \frac{2V}{\sqrt{3g}} = \frac{V}{2} \cdot \frac{2V}{\sqrt{3g}} = \frac{V^2}{\sqrt{3g}}.$

4. The angle $\phi$ with the horizontal can be found using:

$\tan \phi = \frac{y}{x} = \frac{\frac{V^2}{3g}}{\frac{V^2}{\sqrt{3g}}} = \frac{1}{\sqrt{3}} \Rightarrow \phi = \frac{\pi}{6}.$

To determine the direction of the projectile:

1. The vertical velocity $V_y$ can be calculated from:

$V_y = V \sin \theta - gt = \frac{V \sqrt{3}}{2} - g \cdot \frac{2V}{\sqrt{3g}} = \frac{V \sqrt{3}}{2} - \frac{2V}{\sqrt{3}} = V \left(\frac{\sqrt{3}}{2} - \frac{2}{\sqrt{3}}\right).$

2. Simplifying this, we find:

$V \left(\frac{3 - 4}{2\sqrt{3}}\right) = V \left(-\frac{1}{2\sqrt{3}}\right),$
which is negative, thus indicating that the projectile is travelling downwards.

To show this, we start from the acceleration:

1. Since the acceleration is given by:

$\ddot{x} = x - 1,$
we integrate with respect to time to obtain:

2. The velocity equation is:

$\dot{x} = \int (x - 1) dt = xt - \int dt,$
simplifying yields:

$\dot{x} = x - t + C,$
where C is a constant of integration.

3. By solving for C using initial conditions, we will confirm that:

$C = 1$
Hence, we find:

$\dot{x} = x - 1.$

To solve for $x$ as a function of time:

1. Starting from:

$\dot{x} = x - 1,$
we apply separation of variables:

$\frac{dx}{x - 1} = dt.$

2. Integrating both sides gives:

$\ln |x - 1| = t + C.$

3. Solving for $x$, we exponentiate:

$|x - 1| = e^{t + C}.$

4. This leads us to the final expression:

$x(t) = 1 + A e^{t},$
where A depends on initial conditions.

To find the limiting position, we consider the behavior of $x$ as $t \to \infty$:

1. We know:

$x = 1 + A e^{t} \to \infty.$
Hence:

2. We also know that:

$\lim_{t \to \infty} x(t) = \lim_{t \to \infty} (1 + A e^{t}) = \infty.$
Therefore, the limiting position is given as infinity.

To justify this:

1. For player A to win in exactly 7 games, they must win the 5th game on their 7th play.

2. Thus, out of the first 6 games, player A must win exactly 4. This is computed using:

$\binom{6}{4},$
which counts the number of ways to choose 4 games won by player A from the first 6 games.

3. The total number of outcomes of these 7 games, with equal probability of winning, is:

$\left(\frac{1}{2}\right)^7,$
confirming that the probability is:

$P(A) = \binom{7}{5} \left(\frac{1}{2}\right)^7.$

The total probability for player A winning in at most 7 games can be found by summing the probabilities of winning in exactly 5, 6, and 7 games:

1. This gives:

$P(A \text{ wins in at most 7}) = P(A \text{ wins in 5}) + P(A \text{ wins in 6}) + P(A \text{ wins in 7}).$

2. Recognizing the required formulas, we conclude:

$P(A) = \binom{7}{5} \left(\frac{1}{2}\right)^7 + \binom{7}{6} \left(\frac{1}{2}\right)^6 + \binom{7}{7} \left(\frac{1}{2}\right)^5.$

To show this:

1. The scenario involves player A winning a total of (n + 1) games.

2. The number of winning patterns for A is represented by hitting n wins and n losses:

$\binom{2n}{n}$

3. Each game has 2 outcomes (A wins or B wins). Thus, the total number of sequences over(2n) games:

$2^{2n}.$

Hence, it follows that:

$\binom{2n}{n} = 2^{2n}.$