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, we will first determine the vertical component of the velocity when $t = \frac{2V}{\sqrt{3} g}.$ The vertical velocity is given by:

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

Substituting $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and simplifying, we have:

$v_y = \frac{V \sqrt{3}}{2} - \frac{2V}{\sqrt{3}} = \frac{V}{\sqrt{3}} \left(\sqrt{3} - 2\right)$

The horizontal velocity $v_x$ remains constant:

$v_x = V \cos \left(\frac{\pi}{3}\right) = \frac{V}{2}$

Now, the angle $\phi$ with the horizontal is:

$\tan \phi = \frac{v_y}{v_x} = \frac{\frac{V}{\sqrt{3}}(\sqrt{3} - 2)}{\frac{V}{2}} = \frac{2(\sqrt{3} - 2)}{\sqrt{3}}$

To determine the direction of the projectile, we analyze the vertical velocity $v_y$: if $v_y > 0$, it is moving upwards; if $v_y < 0$, it is moving downwards.

From the previous step:

$v_y = \frac{V}{\sqrt{3}}(\sqrt{3} - 2)$

The sign of $v_y$ depends on the value of $\sqrt{3} - 2$:

• Since $\sqrt{3} \approx 1.732$, thus $\sqrt{3} - 2 < 0$, thereby $v_y < 0$.

Consequently, the projectile is travelling downwards at $t = \frac{2V}{\sqrt{3} g}.$

From the given equation, the acceleration is:

$\ddot{x} = x - 1$

Integrating once with respect to time:

$\dot{x} = \int (x - 1) dt$

This gives:

$\dot{x} = \frac{1}{2} e^{-x} + C$

Applying initial conditions: since at $t = 0$, $x(0) = 0$ implies:

$\dot{x} = e^{-0} = 1.$

This results in the integration constant $C = 1$, thus:

$$ \dot{x} = xe^{-x}.$

Starting from:

$\dot{x} = xe^{-x}$

We separate variables:

$\frac{dx}{xe^{-x}} = dt$

Integrating both sides, we find:

$\int \frac{1}{x} dx - \int e^{-x} dx = \int dt$

This leads to:

$\ln |x| + e^{-x} = t + C$

To solve for $x$, we can exponentiate:

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

To find the limiting position, we observe as $t \to \infty$:

$\dot{x} = xe^{-x}$

As $x$ approaches a limiting value, $\dot{x} o 0$, meaning:

$xe^{-x} = 0 \\ \Rightarrow x = 0$.

Thus, the limiting position of the particle is x = 1, since when integrated we find $\lim_{t \to \infty} x(t) = 1.$

In exactly 7 games, player A must win 5 times and player B must win 2 times. The order does not matter, hence we use binomial coefficients. The total number of games is 7, and player A must win the last game (making it 5 wins).

The number of ways player A can win 5 out of the first 6 games (with 2 losses) is given by:

$\binom{6}{2}$.

Each specific outcome occurs with a probability of (\left(\frac{1}{2}\right)^7).

To obtain the prize in at most 7 games, calculate the sum of the probabilities of winning in 5, 6, or 7 games:

$P(A) = P(A \text{ in } 5) + P(A \text{ in } 6) + P(A \text{ in } 7)$

This can be expressed as:

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

Each term corresponds to winning in each of those specific games.

The probability that A gets the prize involves winning $(n + 1)$ total games in $(2n)$ games played.

Using binomials, we combine the number of combinations of outcomes where A wins:

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

Thus, factoring out yields:

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