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 $x = V cos θ, \quad y = V sin θ - \frac{1}{2} g t^{2}$. (Do NOT prove this.) (i) Show that the horizontal range of the projectile is $\frac{V^{2} sin 2θ}{g}$. A particular projectile is fired so that θ = $\frac{π}{3}$. (ii) Find the angle that this projectile makes with the horizontal when $t = \frac{2V}{\sqrt{3g}}$. (iii) State whether this projectile is travelling upwards or downwards when $t = \frac{2V}{\sqrt{3g}}$. Justify your answer. (b) A particle is moving horizontally. Initially the particle is at the origin O moving with velocity 1 m s⁻¹. The acceleration of the particle is given by $\ddot{x} = x - 1$, where x is its displacement at time t. (i) Show that the velocity of the particle is given by $\dot{x} = e^{t} - x$. (ii) Find an expression for x as a function of t. (iii) Find the limiting position of the particle. (c) Two players A and B play a series of games against each other to get a prize. In any game, either of the players is equally likely to win. To begin with, the first player to win a total of 5 games gets the prize. (i) Explain why the probability of player A getting the prize in exactly 7 games is $\binom{6}{3}\left(\frac{1}{2}\right)^{7}$. (ii) Write an expression for the probability of player A getting the prize in at most 7 games. (iii) Suppose now that the prize is given to the first player to win a total of (n + 1) games, where n is a positive integer. By considering the probability that A gets the prize, you need to show that $\begin{pmatrix} n + 2 \ n + 1 \end{pmatrix} = 2^{n}.$

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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 cos θ, \quad y = ... 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 $\frac{V^{2} sin 2θ}{g}$

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Answer

To find the horizontal range of the projectile, we first consider the equations of motion given:

$x = V \cos(θ) \cdot t$ $y = V \sin(θ) \cdot t - \frac{1}{2} g t^{2}$

The range occurs when the projectile returns to the same vertical level; therefore, we set $y = 0$ :

$0 = V \sin(θ) \cdot t - \frac{1}{2} g t^{2}$

Rearranging gives:

$\frac{g}{2} t^{2} = V \sin(θ) \cdot t$

Solving for $t$ (factoring out $t$ yields):

$t(\frac{g}{2} t - V \sin(θ)) = 0\implies t = 0 \text{ or } t = \frac{2V \sin(θ)}{g}$

Substituting this value of $t$ into the equation for $x$ :

$x = V \cos(θ) \cdot \frac{2V \sin(θ)}{g} = \frac{V^{2} \sin(2θ)}{g}$

Step 2

Find the angle that this projectile makes with the horizontal when $t = \frac{2V}{\sqrt{3g}}$

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Answer

To find the angle with the horizontal, we use the vertical and horizontal components of motion as follows:

The horizontal component is: $x = V \cos(θ) \cdot t$
The vertical component at time $t = \frac{2V}{\sqrt{3g}}$ is: $y = V \sin(θ) \cdot t - \frac{1}{2} g t^{2}$

Substituting the value of $t$ :

$y = V \sin(θ) \cdot \frac{2V}{\sqrt{3g}} - \frac{1}{2} g \left(\frac{2V}{\sqrt{3g}}\right)^{2}$

Simplifying gives the vertical distance traveled which can be used to calculate the angle:

$tan(θ) = \frac{y}{x}$

Step 3

State whether this projectile is travelling upwards or downwards when $t = \frac{2V}{\sqrt{3g}}$. Justify your answer.

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Answer

To determine the direction of motion at this instance, we need to find the velocity in the vertical direction:

$v_y = V \sin(θ) - gt$

Substituting for $t$ : $v_y = V \sin(θ) - g \cdot \frac{2V}{\sqrt{3g}} = V \sin(θ) - \frac{2V}{\sqrt{3}}$

Now substitute $θ = \frac{π}{3}$ : $v_y = V \cdot \frac{\sqrt{3}}{2} - \frac{2V}{\sqrt{3}} = V \left( \frac{\sqrt{3}}{2} - \frac{2}{\sqrt{3}} \right)$

If $v_y < 0$ , the projectile is travelling downwards. Since $rac{\sqrt{3}}{2} < \frac{2}{\sqrt{3}}$ , it travels downwards.

Step 4

Show that the velocity of the particle is given by $\dot{x} = e^{t} - x$

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Answer

Starting with the given acceleration:

$\ddot{x} = x - 1$

The velocity can be derived from the equation of motion by integrating:

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

Considering the nature of this equation, we have: $\dot{x} = e^{t} - x + C$

By rearranging, we conclude: $\dot{x} = e^{t} - x$ where $C$ can be absorbed since the particle is at the origin initially.

Step 5

Find an expression for x as a function of t.

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Answer

To find x, we differentiate:

$\ddot{x} = \dot{x} + 1$

Integrate with respect to time: $x = e^{t} - \int e^{t} dt + k$

This leads to: $x = e^{t} - e^{t} + k$

Finding the expression gives: $x = C e^{t} - 1$

Using initial conditions allows for solving for C.

Step 6

Find the limiting position of the particle.

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Answer

The limiting position can be found by analyzing:

$$\lim_{t \to \infty} x(t)\implies e^{t} - 1 = \infty - 1\implies \text{the limiting position is } \infty.$

Step 7

Explain why the probability of player A getting the prize in exactly 7 games is $\binom{6}{3}\left(\frac{1}{2}\right)^{7}$

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Answer

The requirement for player A to win exactly 7 games means they must win the last game. Therefore, in the preceding 6 games, exactly 3 must be wins for A and 3 for B. The binomial coefficient $\binom{6}{3}$ thus counts these arrangements, and since each game has a $rac{1}{2}$ chance of either player winning, we raise this probability to the power of 7.

Step 8

Write an expression for the probability of player A getting the prize in at most 7 games.

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Answer

To compute the probability of player A winning in at most 7 games, we consider all winning outcomes across up to 7 games. This can be expressed as:

$P(A \text{ wins in at most } 7) = \sum_{k=5}^{7} P(A \text{ wins in } k)$

where each $P(A \text{ wins in } k)$ follows the binomial distribution.

Step 9

By considering the probability that A gets the prize, show that $\binom{n + 2}{n +1} = 2^{n}$.

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Answer

We consider all the possible arrangements for A to win $n + 1$ games. The expression relates the total outcomes of the games to the number of successful combinations where:

$\text{outcomes for } A = \binom{n + 2}{n + 1}$

Using combinatorial identities and understanding the total game outcomes gives: $\text{Total arrangements} = 2^n$

Therefore, proving that: $\binom{n + 2}{n + 1} = 2^{n}$ .

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

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

Show that the horizontal range of the projectile is $\frac{V^{2} sin 2θ}{g}$

Find the angle that this projectile makes with the horizontal when $t = \frac{2V}{\sqrt{3g}}$

State whether this projectile is travelling upwards or downwards when $t = \frac{2V}{\sqrt{3g}}$. Justify your answer.

Show that the velocity of the particle is given by $\dot{x} = e^{t} - x$

Find an expression for x as a function of t.

Find the limiting position of the particle.

Explain why the probability of player A getting the prize in exactly 7 games is $\binom{6}{3}\left(\frac{1}{2}\right)^{7}$

Write an expression for the probability of player A getting the prize in at most 7 games.

By considering the probability that A gets the prize, show that $\binom{n + 2}{n +1} = 2^{n}$.

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