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A particle is projected from a point O with speed u m s⁻¹ at an angle α to the horizontal - Leaving Cert Applied Maths - Question 3 - 2021
Question 3
A particle is projected from a point O with speed u m s⁻¹ at an angle α to the horizontal. (i) Show that the range of the particle is \( \frac{u^2 \sin 2\alpha}{g} ... show full transcript
Worked Solution & Example Answer:A particle is projected from a point O with speed u m s⁻¹ at an angle α to the horizontal - Leaving Cert Applied Maths - Question 3 - 2021
Step 1
(i) Show that the range of the particle is \( \frac{u^2 \sin 2\alpha}{g} \)
Answer
To derive the range of a projectile, we start with the equations of motion.
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Time of Flight: The time of flight ( t ) for a projectile is given by: [ t = \frac{2u \sin \alpha}{g} ]
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Horizontal Range: The horizontal range is calculated using: [ \text{Range} = u \cos \alpha \times t ] Substituting the expression for time of flight, we have: [ \text{Range} = u \cos \alpha \times \frac{2u \sin \alpha}{g} = \frac{2u^2 \sin \alpha \cos \alpha}{g} ] Using the identity ( \sin 2\alpha = 2\sin \alpha \cos \alpha ), we find: [ \text{Range} = \frac{u^2 \sin 2\alpha}{g} ]
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Maximum Range at 60°: To find the maximum range, we substitute ( \alpha = 60° ): [ |OQ| = \frac{u^2}{g} ] Therefore, we show that the maximum range occurs when the angle of projection is 60°.
Step 2
(ii) Find the distance \( |PQ| \) in terms of u.
Answer
To find ( |PQ| ):
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Horizontal Distance: For an angle of projection of 60°, we can calculate the range as: [ |OP| = \frac{u^2 \sin 120°}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u^2}{2g} ]
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Vertical Distance: The vertical distance when the angle is 60° is: [ |PQ| = \frac{u^2 \sin 60°}{g} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{g} = \frac{\sqrt{3}u^2}{2g} ]
Hence, we can write:
[ |PQ| = 0.13\frac{u^2}{g} \text{ or } 0.01u^2 \text{ when rounded} ]
Step 3
(i) Find d in terms of u.
Answer
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Initial Velocity Components: The initial velocity components along the inclined plane can be decomposed: [ u_x = u \cos 30° \quad and \quad u_y = u \sin 30° ]
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Time of Flight on Incline: The total time of flight can be given as: [ t = \frac{2u \sin 30°}{g} \quad or \quad t = \frac{u}{3g} ]
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Calculating d: The range along the inclined plane is: [ d = u \cos 30° \times t = u \cdot \frac{\sqrt{3}}{2} \times \frac{u}{3g} = \frac{u^2 \sqrt{3}}{6g} ]
Step 4
(ii) Find the value of e.
Answer
To calculate the coefficient of restitution ( e ):
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Velocity Upon Impact: The velocity components before it strikes the inclined plane: [ v_y = u \sin 30° - gt = u \cdot \frac{1}{2} - g \cdot \frac{u}{3g} = \frac{u}{2} - \frac{u}{3} = \frac{u}{6} ] The rebound velocity along the y-axis: [ v_{rebound} = -\left( e \cdot v_y \right) ]
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Using Tan Rule: At an angle of 60° with the horizontal: [ an 60° = \frac{\frac{u}{6}}{\frac{u}{3}} = e \Rightarrow e = 1 ] Therefore, the coefficient of restitution ( e ) is confirmed as 1.