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 ( \phi ) at a given time, we use the vertical velocity component:

1. The vertical velocity ( v_y ) is given by: $v_y = V \sin \theta - g t$

2. Substitute ( \theta = \frac{ \pi}{3} ) and ( t = \frac{2V}{\sqrt{3}g} ): $v_y = V \sin \left(\frac{\pi}{3}\right) - g \left(\frac{2V}{\sqrt{3}g}\right)$ $= \frac{\sqrt{3}}{2}V - \frac{2V}{\sqrt{3}}$ Now, calculate ( v_y ).

3. The angle ( \phi ) can then be found using: $\tan \phi = \frac{v_y}{v_x}$ where ( v_x = V \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}V $$. Thus solve for ( \phi ).

To determine if the projectile is traveling upwards or downwards at ( t = \frac{2V}{\sqrt{3}g} ), we examine the sign of the vertical component of velocity ( v_y ):

1. As established earlier, if ( v_y > 0 ), the projectile is moving upwards; if ( v_y < 0 ), it is moving downwards.
2. Calculate ( v_y ): If the calculated value of ( v_y ) yields a negative result, it indicates that the projectile is traveling downwards.

Thus, based on the value of ( v_y ), we can conclude the direction of motion.

To derive the velocity of the particle:

1. We start with the acceleration equation: $\ddot{x} = x - 1$ which can be rewritten as: $\ddot{x} = \frac{dx}{dt}$.

2. Integrate this equation with respect to time: $\dot{x} = \int (x - 1) dt$ This results in: $\dot{x} = e^t - x$, where we set initial conditions to solve for constants.

From the previous step, where we found ( \dot{x} = e^t - x ), we can separate variables:

1. Rearranging gives: $dx = (e^t - x) dt$

2. This is a separable differential equation. Rearranging and integrating, $\int \frac{dx}{e^t - x} = \int dt$.

3. Solve to find the expression for x in terms of t, applying relevant integration techniques.

The limiting position occurs when the time approaches infinity. Thus, take the limit:

1. From the earlier derived function using the relation: $\lim_{t \to \infty} x(t)$

2. Set ( e^t ) as dominant, and integrate accordingly to find: The limiting position is the steady-state value to which the particle converges.

To win the prize in exactly 7 games, player A must win the 7th game:

1. Player A must win 5 games, while player B wins 2 games. The arrangement of victories is crucial.
2. Player A can win any 5 of the 6 games played before the 7th game.
3. Thus, the total arrangements is given by: ( \binom{6}{1} ) since B winning includes only winning 2 of the games in the first 6.
4. The probability for each game is ( \left(\frac{1}{2}\right)^7 ), since each game is independent.

The probability of player A winning the prize in at most 7 games can be determined by considering outcomes from 5 to 7 games. Hence:

1. Use the binomial theorem and sum up winning probabilities for each outcome: Let ( P(A) ) be represented as: $P(A) = \sum_{k=5}^{7} P(A \text{ wins in } k \text{ games})$
2. Expand using earlier derived probabilities for 5, 6 and 7 games.

To derive this, follow these steps:

1. Start by calculating all combinations of games with player A winning a total of (n + 1) games against B under the condition that A needs one additional win.
2. Use combinations and factorial arrangements to express total outcomes: $P(A) = \binom{n + 1}{n}$
3. Finally, apply the binomial probability theorem to arrive at the stated equality, simplifying as necessary to prove equality.