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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
Question 14
A projectile is fired from the origin O with initial velocity V m s⁻¹ at an angle θ to the horizontal. The equations of motion are given by $$x = V \, ext{cos} \th... show full transcript
Worked Solution & Example Answer: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
Step 1
Show that the horizontal range of the projectile is \( \frac{V^2 \sin 2\theta}{g} \).
Answer
To find the horizontal range ( R ) of the projectile, we use the equations of motion:
- The time of flight ( T ) can be found by setting ( y = 0 ), which gives ( T = \frac{2V \sin \theta}{g} ).
- The horizontal range is determined by ( R = x(T) = V \cos \theta \cdot T = V \cos \theta \cdot \frac{2V \sin \theta}{g} = \frac{2V^2 \sin \theta \cos \theta}{g} = \frac{V^2 \sin 2 \theta}{g} ) using the identity ( \sin 2\theta = 2 \sin \theta \cos \theta ).
Step 2
Find the angle that this projectile makes with the horizontal when \( t = \frac{2V}{\sqrt{3}g} \).
Answer
At time ( t = \frac{2V}{\sqrt{3}g} ), we can find the vertical position:
- Using the equation for vertical motion, ( y ), we substitute ( t ) to get:
- Simplifying, we calculate the angle ( \beta ) using ( \tan \beta = \frac{y}{x} ). We find the respective angle using arctangent.
Step 3
State whether this projectile is travelling upwards or downwards when \( t = \frac{2V}{\sqrt{3}g} \). Justify your answer.
Answer
To determine whether the projectile is travelling upwards or downwards, we calculate the vertical velocity using:
- Substituting ( t = \frac{2V}{\sqrt{3}g} ) and ( \theta = \frac{\pi}{3} ), we conclude:
- If ( v_y > 0 ), the projectile is travelling upwards; if ( v_y < 0 ), it is travelling downwards. Therefore, we compute to see the sign of ( v_y ).
Step 4
Show that the velocity of the particle is given by \( \dot{x} = e^{t} - x. \)
Answer
Starting from the acceleration relation ( \ddot{x} = x - 1 ), we can express:
- Integrating ( \ddot{x} = x - 1 ) gives us a first-order differential equation.
- This leads us to solve for ( \dot{x} ) in terms of time by integrating, yielding:
Step 5
Find an expression for x as a function of t.
Answer
To find ( x(t) ), we recognize that this first-order equation can be solved using an integrating factor or separation of variables:
- Solve the form( e^{t} dx + x dt = 0 ) or via Laplace transform where necessary.
- Ultimately, we express ( x(t) ) in terms of the initial conditions and the manipulated equation.
Step 6
Find the limiting position of the particle.
Answer
The limiting position is found by checking the behavior as ( t \to \infty ):
- By evaluating ( \lim_{t \to \infty} \dot{x} , ) and setting it to steady velocity terms, we find the limit to guide us to:
- Thus, the limiting position is established at that stage.
Step 7
Explain why the probability of player A getting the prize in exactly 7 games is \( \binom{6}{3} \left( \frac{1}{2} \right)^7 \).
Answer
For player A to win in exactly 7 games:
- They must win 4 games and B must win 3, which makes for the configuration counted by ( \binom{6}{3} ): 3 wins must occur in the first 6 games.
- Thus the formula represents the total arrangements multiplied by the probability of overall wins.
Step 8
Write an expression for the probability of player A getting the prize in at most 7 games.
Answer
The probability of player A winning in at most 7 games includes scenarios where they can win 4, 5, or 6 games in this series:
- Therefore, we compute the total probability from win configurations calculated earlier retaining up to the 7th game.
Step 9
By considering the probability that A gets the prize, show that \( P(A) = \binom{2n}{n} \cdot \frac{1}{2^{2n}}. \)
Answer
Using combinatorial arguments and corresponding games won scenarios help reach towards this conclusion:
- Here, we analyze winning sequences under alternating wins, articulating those arrangements to achieve ( P(A) = \binom{2n}{n} \frac{1}{2^{2n}}. )