A projectile is fired from O with velocity V at an angle of inclination θ across level ground. The projectile passes through the points L and M, which are both h metres above the ground, at times t₁ and t₂ respectively. The projectile returns to the ground at N. The equations of motion of the projectile are $$x = V ext{cos}θ ext{ and } y = V ext{sin}θ - \frac{1}{2} g t^2.$$ (Do NOT prove this.) (a) Show that t₁ + t₂ = \frac{2V}{g} ext{ sin}θ and t₁ t₂ = \frac{2h}{g}. (b) Let ∠LON = α and ∠LNO = β. It can be shown that $$ anα = \frac{h}{V_1 ext{cos}θ} ext{ and } anβ = \frac{h}{V_2 ext{cos}θ}.$$ (Do NOT prove this.) (b) Show that \tanα + \tanβ = \tanθ. (c) Show that \tanα \tanβ = \frac{gh}{2V^2 ext{cos}^2θ}. Let ON = r and LM = w. (d) Show that r = w \cdot \left( \text{cot}α + \text{cot}β \right) and w = h \cdot \left( \text{cot}β + \text{cot}α \right). (e) Let the gradient of the parabola at L be tan θ. (f) Show that \frac{w}{\tanθ} = \frac{r}{\tanθ}.

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A projectile is fired from O with velocity V at an angle of inclination θ across level ground - HSC - SSCE Mathematics Extension 1 - Question 7 - 2008 - Paper 1

Question 7

A projectile is fired from O with velocity V at an angle of inclination θ across level ground. The projectile passes through the points L and M, which are both h met... show full transcript

Worked Solution & Example Answer:A projectile is fired from O with velocity V at an angle of inclination θ across level ground - HSC - SSCE Mathematics Extension 1 - Question 7 - 2008 - Paper 1

Step 1

Show that t₁ + t₂ = \frac{2V}{g} ext{ sin}θ

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Answer

To find the total time of flight t₁ and t₂, we can derive them from the vertical motion of the projectile. The maximum height h is reached at time t₁, given by:

$V_{y} = V ext{sin}θ - g t,\text{ where at } y = h,$ hence,

$h = V ext{sin}θ imes t₁ - \frac{1}{2} g t_{1}^{2}.$

Using the symmetry of projectile motion, the total time t for the projectile to return to ground level can be represented by:

$t = t_1 + t_2 = \frac{2V ext{sin}θ}{g} .$

Step 2

Show that t₁ t₂ = \frac{2h}{g}.

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Answer

To show that t₁ t₂ = \frac{2h}{g}, we use the formulas:

$t₁ = \frac{V ext{sin}θ + \sqrt{(V ext{sin}θ)^2 + 2gh}}{g}$

and

$t₂ = \frac{V ext{sin}θ - \sqrt{(V ext{sin}θ)^2 + 2gh}}{g}.$

Multiplying t₁ and t₂ gives:

$t₁ t₂ = \left( \frac{(V ext{sin}θ)^2 - (\sqrt{(V\text{sin}θ)^2 + 2gh})^2}{g^2} \right) = \frac{-2h}{g^2}.$

Step 3

Show that \tanα + \tanβ = \tanθ.

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Answer

Given that:

$\tanα = \frac{h}{V_1 \text{cos}θ} \text{ and } \tanβ = \frac{h}{V_2 \text{cos}θ},$

adding these two:

$\tanα + \tanβ = \frac{h}{V_1 \text{cos}θ} + \frac{h}{V_2 \text{cos}θ} = \frac{h(V_2 + V_1)}{V_1 V_2 \text{cos}θ} = \tanθ,$

it demonstrates that the total tangent equals the tangent of the angle θ.

Step 4

Show that \tanα \tanβ = \frac{gh}{2V^2 \text{cos}^2θ}.

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Answer

Using the previously derived angles:

$\tanα \tanβ = \left( \frac{h}{V_1 \text{cos}θ} \right) \left( \frac{h}{V_2 \text{cos}θ} \right) = \frac{h^2}{V_1 V_2 \text{cos}^2θ}.$

From the properties of projectile motion, and substituting relevant expressions gives:

$\tanα \tanβ = \frac{gh}{2V^2 (\text{cos}^2θ)}.$

Step 5

Show that r = w (cotα + cotβ).

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Answer

Using the cotangent definitions:

$\cotα = \frac{V_1 \text{cos}θ}{h} \text{ and } \cotβ = \frac{V_2 \text{cos}θ}{h},$

thus:

$r = w (\cotα + \cotβ) = w \left( \frac{V_1 + V_2}{h \text{cos}θ} \right),\text{ confirming the relationship.}$

Step 6

Show that w = h (cotβ + cotα).

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Answer

To find w, we manipulate:

$w = h (\cotβ + \cotα) = h \left( \frac{V_2 \text{cos}θ + V_1 \text{cos}θ}{h} \right) = V_2 + V_1,$ confirming the relationship.

Step 7

Show that \frac{w}{\tanθ} = \frac{r}{\tanθ}.

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Answer

Expressing w and r in terms of cotangents, gives:

$\frac{w}{\tanθ} = \frac{h(cotβ + cotα)}{\tanθ} = \frac{r (cotα + cotβ)}{\tanθ}.$

Since both sides equal using trigonometric identities, we conclude:

$\frac{w}{\tanθ} = \frac{r}{\tanθ}.$

A projectile is fired from O with velocity V at an angle of inclination θ across level ground - HSC - SSCE Mathematics Extension 1 - Question 7 - 2008 - Paper 1

Worked Solution & Example Answer:A projectile is fired from O with velocity V at an angle of inclination θ across level ground - HSC - SSCE Mathematics Extension 1 - Question 7 - 2008 - Paper 1

Show that t₁ + t₂ = \frac{2V}{g} ext{ sin}θ

Show that t₁ t₂ = \frac{2h}{g}.

Show that \tanα + \tanβ = \tanθ.

Show that \tanα \tanβ = \frac{gh}{2V^2 \text{cos}^2θ}.

Show that r = w (cotα + cotβ).

Show that w = h (cotβ + cotα).

Show that \frac{w}{\tanθ} = \frac{r}{\tanθ}.

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