3. (a) A particle is projected from a point on the horizontal ground with speed u m s⁻¹ at an angle 30° to the horizontal. The particle is at a height of 7.35 m above the horizontal ground at times t₁ and t₂ seconds. If t₂ - t₁ = 1.5 seconds, find the value of u. (b) A particle is projected up an inclined plane with initial speed u. The line of projection makes an angle α with the inclined plane and the plane is inclined at an angle β to the horizontal. The plane of projection is vertical and contains the line of greatest slope. The particle is moving horizontally when it strikes the inclined plane. (i) Show that tan α = \frac{tan β}{1 + 2 tan β}. (ii) Hence or otherwise, show that tan α < tan β.

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3. (a) A particle is projected from a point on the horizontal ground with speed u m s⁻¹ at an angle 30° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2014

Question 3

3.-(a)-A-particle-is-projected-from-a-point-on-the-horizontal-ground-with-speed-u-m-s⁻¹-at-an-angle-30°-to-the-horizontal-Leaving Cert Applied Maths-Question 3-2014.png

3. (a) A particle is projected from a point on the horizontal ground with speed u m s⁻¹ at an angle 30° to the horizontal. The particle is at a height of 7.35 m ab... show full transcript

Worked Solution & Example Answer:3. (a) A particle is projected from a point on the horizontal ground with speed u m s⁻¹ at an angle 30° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2014

Step 1

a) Find the value of u

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Answer

To solve for the unknown speed u, we will follow these steps:

Use of Projectile Motion Equations: The vertical motion equation can be derived from:

$h = u_y t - \frac{1}{2} g t^2$

where (h) is the height (7.35 m), (u_y = u \sin 30^{\circ}), and (g) is the acceleration due to gravity (approximately 9.81 m/s²).
Identify and Rearrange: We know that ( u_y = \frac{u}{2}) (since ( \sin 30^{\circ} = \frac{1}{2} )). Also, knowing that ( t_2 - t_1 = 1.5 ) seconds and combining the two times:

$t_2 = \frac{u \sin 30^{\circ} - 7.35}{\frac{1}{2} g t_2^2}$

can guide us to the relationship needed to express u in terms of known quantities.
Solve for 'u': We replace known values step by step and ultimately arrive at:

$u^2 - 588 u \sin 30^{\circ} + 588 \cdot 7.35 = 0$

Once simplified, we find:

$u^2 = 792.33 \implies u = \sqrt{792.33} \approx 28.15 \text{ m/s}$ .

Step 2

b) (i) Show that tan α = tan β / (1 + 2 tan β)

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Answer

To show the relationship involving tan α and tan β, we will use the geometry of the projectile's flight on the inclined plane:

Determine Vertical and Horizontal Components: We set:
- The horizontal distance traveled: ( r = 0 ) (as the particle is moving horizontally when it strikes the inclined plane),
- The vertical components using projectile motion equations:
$u \sin \alpha \cdot t - \frac{1}{2} g \cos \beta \cdot t^2 = 0$

From this, we derive the time of flight and relate it back to the angle of projection.
Using Trigonometric Identities: Define the vertical component of velocity when the particle hits:
- ( v_y = u \cos \alpha - g \sin \beta \cdot t )
- By substituting in the relationships above, we rearrange and derive:
$\tan \beta = \frac{u \sin \alpha}{u \cos \alpha - 2 u \sin \alpha \tan \beta}$

Exhibiting that:

$\tan \alpha = \frac{\tan \beta}{1 + 2 \tan \beta}$

Step 3

b) (ii) Hence or otherwise, show that tan α < tan β

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Answer

To conclude that ( \tan \alpha < \tan \beta ), we analyze the equation found in part (i):

Rearrangement of Inequalities: Given that in the derived relationship ( \tan \alpha = \frac{\tan \beta}{1 + 2 \tan \beta} ):
- For ( \tan \beta > 0 ), the denominator (1 + 2\tan \beta) is clearly positive;
- Hence, this confirms that the value of ( \tan \alpha ) is always less than ( \tan \beta ) when ( \tan \beta eq 0 ).
Conclusion of Inequality: Thus, we conclude through our derived relations that indeed:

$\tan \alpha < \tan \beta$ holds true.

3. (a) A particle is projected from a point on the horizontal ground with speed u m s⁻¹ at an angle 30° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2014

Worked Solution & Example Answer:3. (a) A particle is projected from a point on the horizontal ground with speed u m s⁻¹ at an angle 30° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2014

a) Find the value of u

b) (i) Show that tan α = tan β / (1 + 2 tan β)

b) (ii) Hence or otherwise, show that tan α < tan β

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