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

Using the vertical motion equation:

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

Substituting (t = \frac{2V}{\sqrt{3}g}):

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

This simplifies to:

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

The horizontal velocity is (V \cos(\frac{\pi}{3}) = \frac{V}{2}\

Using (\tan(θ) = \frac{y}{x}), we can find (θ).

At time (t = \frac{2V}{\sqrt{3}g}), we can find the vertical velocity (V_y = V \sin(\frac{\pi}{3}) - g t). If this value is positive, the projectile is travelling upwards; otherwise, it travels downwards.

Calculating gives:

Given the equation of motion (\ddot{x} = x - 1), we integrate to find the velocity: (\dot{x} = \int (x - 1) dt).

This leads to:

$$\dot{x} = e^{t - x}.$

Rearranging and solving (\dot{x} = e^{t - x}) gives a separable differential equation. Integrate accordingly to find:

$x(t) = f(t) + C$ where C is a constant determined by the initial conditions.

The limiting position is reached as (t \to \infty), where (x(t)) approaches a constant value derived from solving the limit of the integrated function.

Player A must win exactly 4 games out of 6 prior games and then win the 7th game. Thus, the selection is represented by (\binom{6}{1}). Multiplying by the probabilities yields: (\left(\frac{1}{2}\right)^{7}).

The probability can be expressed as the sum of probabilities for winning in each game count scenario from 5 to 7, showing a cumulative distribution function representation.

This expression arises from combinatorial analysis evaluating the pathway and permutations of the game's outcomes leading to player A's victory, as derived from the total win conditions evaluated in the probability landscape.