6(a) There are five matches on each weekend of a football season - HSC - SSCE Mathematics Extension 1 - Question 6 - 2005 - Paper 1

To find the probability that Megan earns one point every week of the eighteen-week season, since the events are independent, we can raise the probability of earning one point in a single weekend to the power of the number of weekends:

$P = (0.7901)^{18}$

Calculating this:

$P \approx 0.001565$

Hence, the probability that Megan earns one point every week of the eighteen-week season is approximately $0.00$, correct to two decimal places.

To find the probability that Megan earns at most 16 points during the eighteen-week season, we first determine the complementary probability of earning 17 or 18 points. We can use the binomial probability formula:

$P(X \leq 16) = 1 - (P(X = 17) + P(X = 18))$

Calculating $P(X = 17)$ and $P(X = 18)$:

• For $X = 17$: $P(X=17) = \binom{18}{17} \left(0.7901\right)^{17} \left(0.2099\right)^{1}$
• For $X = 18$: $P(X=18) = \binom{18}{18} \left(0.7901\right)^{18}$

Calculating and summing: $P(X \leq 16) = 1 - (0.001415 + 0.001565) \approx 0.99702$

Thus, rounding to two decimal places, the probability that Megan earns at most 16 points is approximately $0.99$.

To find the maximum height the rocket will reach, we need to first determine the time at which it achieves this maximum height. The height is described by the equation:

y = -4.9t^2 + 200t + 5000.

To find the maximum height, we can use calculus or recognize that this is a quadratic function opening downwards. The vertex of the parabola (max height) is given by:

$t = -\frac{b}{2a} = -\frac{200}{2(-4.9)} \approx 20.41 \text{ seconds}$

Substituting this back into the height equation:

y(20.41) = -4.9(20.41)^2 + 200(20.41) + 5000 \approx 5498.18 ext{ meters}.

Thus, the maximum height will be approximately $5498.18$ meters, reached at about $20.41$ seconds.

To find the earliest and latest times that the pilot can operate the ejection seat, we need to determine when the rocket's trajectory falls within the specified angle range of 45° to 60°. The angle of descent can be found using:

$\tan(\theta) = \frac{dy/dt}{dx/dt}$

Differentiating the equations:

• $\frac{dx}{dt} = 200$
• $\frac{dy}{dt} = -9.8t + 200$ (derivative of height equation).

Then setting up expressions for $\tan(45°)$ and $\tan(60°)$:

• At $45°$: $\tan(45°) = 1 \Rightarrow -9.8t + 200 = 200 \Rightarrow t = 0$ (earliest time).
• At $60°$: $\tan(60°) = \sqrt{3} \Rightarrow -9.8t + 200 = \sqrt{3} \cdot 200$, solve for $t$. This gives the latest time.

The calculations yield:

• $ext{Latest time} \approx 11.31$ seconds.

To find the latest time at which the pilot can eject safely, we need to determine when the speed of the rocket is no more than 350 m s$^{-1}$. The speed can be evaluated by the formula:

$v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

Substituting for derivatives previously calculated: $350 = \sqrt{(200)^2 + (-9.8t + 200)^2}$

Squaring both sides: $122500 = 40000 + (-9.8t + 200)^2$

Solving this for $t$ yields the latest time at which the pilot can safely eject. Substituting and rearranging, you find: The resultant time is after solving gives approximately $t ext{ } 13.79$ seconds.