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

Using the vertical motion equation:

$y = V \sin(\frac{\pi}{3}) t - \frac{1}{2} g t^2$

We substitute $\theta = \frac{\pi}{3}$ and $t = \frac{2V}{\sqrt{3g}}$:

$y = \frac{V \sqrt{3}}{2} \cdot \frac{2V}{\sqrt{3g}} - \frac{1}{2} g \left( \frac{2V}{\sqrt{3g}} \right)^2$

Calculating the two terms:

1. The first term: $= \frac{V^2}{g}$

2. The second term: $= \frac{2V^2}{3g}$

Thus we find:

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

To find the angle with the horizontal, use:
$\tan(\theta_{angle}) = \frac{y}{x}.$ We find $x$ using: $x = V \cos(\frac{\pi}{3}) \cdot \frac{2V}{\sqrt{3g}} = \frac{V^2}{\sqrt{3g}}.$

Now, substituting back: $\tan(\theta_{angle}) = \frac{\frac{V^2}{3g}}{\frac{V^2}{\sqrt{3g}}} = \frac{1}{\sqrt{3}}.$

Thus, \theta_{angle} = \frac{\pi}{6}.

To determine whether the projectile is moving upwards or downwards at time $t = \frac{2V}{\sqrt{3g}}$, we examine its vertical velocity $v_y$:

$v_y = V \sin(\theta) - g t.$

Substituting in our known values: $v_y = V \cdot \frac{\sqrt{3}}{2} - g \cdot \frac{2V}{\sqrt{3g}} = \frac{V \sqrt{3}}{2} - \frac{2V}{\sqrt{3}}.$

By simplifying, we find: $v_y = V \left( \frac{\sqrt{3}}{2} - \frac{2}{\sqrt{3}} \right).$

This shows a negative result, thus indicating that the projectile is travelling downwards at this moment in time.

Given the acceleration equation:

$\ddot{x} = \dot{x} \cdot \frac{dx}{dt} = x - 1,$

we can integrate both sides according to the standard method. By rearranging and substituting, we find:

To integrate, we separate variables: $\frac{dx}{x - 1} = dt.$

Integrating yields: $\ln|x - 1| = t + C,$
where $C$ is the constant of integration. Exponentiating both sides leads to: $x - 1 = e^{t + C}.$

Thus, the expression for velocity becomes: $\dot{x} = e^{t - x}.$

From the previous part, we have:

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

Solving for $x$ gives: $x = e^{t + C} + 1.$

Now we determine the constant $C$ using the initial condition: At $t=0$, $x=0$, hence:

$0 - 1 = e^{0 + C} \Rightarrow C = -\ln(1) \Rightarrow C=0.$

Thus, we replace to find:

$x = e^{t} + 1.$

The limiting position is found as $t \to \infty$:

$x = e^{t} + 1 \to \infty + 1,$

which implies the limiting position is:

$\infty.$

Player A must win exactly 4 games and must lose exactly 3 games within 7 games in total, leading to:

1. The win conditions: Player A wins the 4th game, hence there are 6 prior games where A wins 3 and B wins 3, which can be arranged in:

$\binom{6}{3}.$

2. The overall probability of each configuration occurring, since the games are independent:

$\left( \frac{1}{2} \right)^{7}.$

The probability that player A wins in at most 7 games can be expressed as:

$\sum_{k=4}^{5} \binom{6}{k-1} \left( \frac{1}{2} \right)^{7}.$

This sum denotes the probabilities of obtaining 4 or 5 wins in 7 games, respectively.

To demonstrate that:

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

consider the scenarios where player A wins 2n games total, thus every configuration can be represented through combinations of wins and losses. By the binomial theorem, this leads to:

$(\left( \frac{1}{2} + \frac{1}{2} \right)^{2n} = 2^{2n}.$

Thus proving that the required relationship holds true.