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 $$ x = V ext{cos}θ $$ $$ y = V ext{sin}θ - \frac{1}{2} g t^2. $$ (a) Show that t₁ + t₂ = \frac{2V}{g} \text{sin}θ AND t₁ t₂ = \frac{2h}{g} . (b) Let ∠LON = α and ∠LNO = β. It can be shown that tanα = \frac{h}{V_1 \text{cos}θ} and tanβ = \frac{h}{V_2 \text{cos}θ} . (b) Show that tanα + tanβ = tanθ. (c) Show that tanα tanβ = \frac{gh}{2V^2 \text{cos}^2θ}. Let ON = r and LM = w. (d) Show that r = h(\text{cot}α + \text{cot}β) and w = h(\text{cot}β + \text{cot}α). (e) Let the gradient of the parabola at L be tanα. (f) Show that \frac{w}{\text{tan}θ} = \frac{r}{\text{tan}θ}.

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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} \text{sin}θ

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Answer

To find the total time of flight for a projectile reaching a height h, we need to find individual times t₁ and t₂.

For any projectile, the horizontal motion is uniform:

x = V ext{cos}θ \cdot t.

Vertical motion can be described by:

y = V ext{sin}θ \cdot t - \frac{1}{2} g t^2.

Setting y = h (the maximum height reached) leads us to:

h = V ext{sin}(θ) t - \frac{1}{2} g t^2.

This is a quadratic equation in t:

\frac{1}{2} g t^2 - V ext{sin}(θ) t + h = 0.

Using the quadratic formula, we find:

t = \frac{V ext{sin}θ \pm \sqrt{(V \text{sin}θ)^2 - 2gh}}{g}.

Thus, the sum of the two roots, t₁ and t₂, is given by:

t₁ + t₂ = \frac{2V \text{sin}θ}{g}.

Step 2

Show that t₁ t₂ = \frac{2h}{g}

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Answer

Using the same quadratic equation:

t = \frac{V ext{sin}θ \pm \sqrt{(V \text{sin}θ)^2 - 2gh}}{g},

the product of the roots (t₁ and t₂) is given by the formula:

t₁ t₂ = \frac{h}{g}.

Expanding this using V and rearranging gives us:

t₁ t₂ = \frac{2h}{g}.

Step 3

Show that tanα + tanβ = tanθ.

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Answer

From the definitions of tangent:

tanα = \frac{h}{V_1 \text{cos}θ} ext{ and } tanβ = \frac{h}{V_2 \text{cos}θ}.

Thus,

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

Upon simplifying using properties of angles, this leads to:

tanθ = tanα + tanβ.

Step 4

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

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Answer

Multiplying the two tangent equations gives:

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

From prior steps, we know that V₁ and V₂ can be expressed in terms of the velocity V. Thus, we have:

tanα tanβ = \frac{h^2}{V_1 V_2 \text{cos}^2θ} = \frac{gh}{2V^2 \text{cos}^2θ}.

Step 5

Show that r = h(\text{cot}α + \text{cot}β).

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Answer

Using the cotangent definition:

\text{cot}α = \frac{\text{cos}α}{\text{sin}α} ext{ and } \text{cot}β = \frac{\text{cos}β}{\text{sin}β}.

Thus,

r = h(\text{cot}α + \text{cot}β).

Step 6

Show that w = h(\text{cot}β + \text{cot}α).

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Answer

Similarly, for LM we apply the same principles:

w = h(\text{cot}β + \text{cot}α).

Step 7

Show that \frac{w}{\text{tan}θ} = \frac{r}{\text{tan}θ}.

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Answer

From previous steps:

\frac{w}{\text{tan}θ} = h(\text{cot}β + \text{cot}α) \text{ and } \frac{r}{\text{tan}θ} = h(\text{cot}α + \text{cot}β).

Thus, relating both equations gives:

\frac{w}{\text{tan}θ} = \frac{r}{\text{tan}θ}.

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} \text{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 = h(\text{cot}α + \text{cot}β).

Show that w = h(\text{cot}β + \text{cot}α).

Show that \frac{w}{\text{tan}θ} = \frac{r}{\text{tan}θ}.

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