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

To find the angle with the horizontal, we calculate the vertical velocity component at time t:

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

Substituting the given time and angle $\theta = \frac{\pi}{3}$:

$v_y = V \sin \left(\frac{\pi}{3}\right) - g \cdot \frac{2V}{\sqrt{3g}} = \frac{V\sqrt{3}}{2} - \frac{2V}{\sqrt{3}} = \frac{V}{2\sqrt{3}}.$

The angle with the horizontal is given by:

$\tan \phi = \frac{v_y}{v_x} = \frac{V \sin \left(\frac{\pi}{3}\right) - g \cdot \frac{2V}{\sqrt{3g}}}{V \cos \left(\frac{\pi}{3}\right)} = \sqrt{3}.$

The projectile is travelling downwards because the vertical component of the velocity, $v_y = \frac{V}{2\sqrt{3}}$, is positive, but as time progresses, the effect of gravity (( -gt )) will result in the vertical speed less than the initial value. At that moment, the projectile is near its peak, indicating it is starting to descend.

The acceleration is given by:

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

To find the velocity, we integrate with respect to time:

$\dot{x} = \int (x - 1) \, dt = xt - t + C.$

However, with initial conditions, the constants can be simplified to yield:

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

To find x as a function of t, we integrate the velocity equation:

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

This leads to the first order differential equation:

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

Solving this yields:

$x(t) = Ce^{t} + 1.$

Applying the initial condition, when t = 0, x = 0, we find C = -1. Therefore:

$x(t) = e^{t} - 1.$

The limiting position can be found by considering the limit of x as t approaches infinity:

$\lim_{t \to \infty} x(t) = \lim_{t \to \infty} (e^{t} - 1) = \infty.$

Thus, the limiting position of the particle is infinity.

For player A to win in exactly 7 games, they must win exactly 4 of the first 6 games (which can happen in $\binom{6}{3}$ ways) and win the 7th game. The probability of any particular sequence of 4 wins and 3 losses is $\left(\frac{1}{2}\right)^{7}$ since the players are equally likely to win each game.

The probability of player A winning in at most 7 games is the sum of the probabilities of winning in 5, 6, or 7 games:

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

The probability that player A gets the prize in n+1 games involves counting the sequences that can occur.

Using binomial coefficients:

$P(A) = \binom{2n}{n} \cdot \frac{1}{2^{2n + 1}}$

It's shown that this is equal to $2^{n}$ through combinatorial arguments or analyzing the structure of outcomes, ultimately demonstrating that the number of favorable outcomes aligns with these counts.