A particle is projected from a point P, as shown in the diagram, with an initial speed of 74 m s⁻¹ at an angle β to the horizontal, where $ \tan \beta = \frac{1}{2} $. The particle reaches point Q after 4 seconds of motion. R is the highest point reached by the particle. Find (i) the initial velocity of the particle in terms of $ \hat{i} $ and $ \hat{j} $ (ii) the velocity of the particle at point Q in terms of $ \hat{i} $ and $ \hat{j} $ (iii) the displacement of R from P in terms of $ \hat{i} $ and $ \hat{j} $ (iv) the value of k, given that the particle reaches S after 16 seconds of motion.

Leaving Cert Applied Maths Projectiles: A particle is projected from a p

A particle is projected from a point P, as shown in the diagram, with an initial speed of 74 m s⁻¹ at an angle β to the horizontal, where $ \tan \beta = \frac{1}{2} $ - Leaving Cert Applied Maths - Question 3 - 2016

Question 3

$A-particle-is-projected-from-a-point-P,-as-shown-in-the-diagram,-with-an-initial-speed-of-74-m-s⁻¹-at-an-angle-β-to-the-horizontal,-where-$-\tan-\beta-=-\frac{1}{2}-$-Leaving Cert Applied Maths-Question 3-2016.png$

A particle is projected from a point P, as shown in the diagram, with an initial speed of 74 m s⁻¹ at an angle β to the horizontal, where \( \tan \beta = \frac{1}{2}... show full transcript

Worked Solution & Example Answer:A particle is projected from a point P, as shown in the diagram, with an initial speed of 74 m s⁻¹ at an angle β to the horizontal, where $ \tan \beta = \frac{1}{2} $ - Leaving Cert Applied Maths - Question 3 - 2016

Step 1

(i) the initial velocity of the particle in terms of $ \hat{i} $ and $ \hat{j} $

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Answer

To find the initial velocity ( \mathbf{u} ), we can decompose the initial speed into its horizontal and vertical components.
[ \mathbf{u} = 74\cos(\beta) \hat{i} + 74\sin(\beta) \hat{j} ]
Given that ( \tan(\beta) = \frac{1}{2} ), we can write:
( \sin(\beta) = \frac{1}{\sqrt{5}} ) and ( \cos(\beta) = \frac{2}{\sqrt{5}} ).
Therefore,
[ \mathbf{u} = 74 \left(\frac{2}{\sqrt{5}}\right) \hat{i} + 74 \left(\frac{1}{\sqrt{5}}\right) \hat{j} ]
[ \mathbf{u} = 24\hat{i} + 74\hat{j} ]

Step 2

(ii) the velocity of the particle at point Q in terms of $ \hat{i} $ and $ \hat{j} $

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Answer

At point Q, the time taken is 4 seconds. We will use the following equations of motion to find the velocity:
[ \mathbf{v} = \mathbf{u} + \mathbf{a}t ]
where the acceleration ( \mathbf{a} = -g \hat{j} ).
Then,
[ \mathbf{v} = \left(24\hat{i} + 74\hat{j}\right) + \left(0\hat{i} - 10\hat{j}\right) ]
[ \mathbf{v} = 24\hat{i} + 30\hat{j} ]

Step 3

(iii) the displacement of R from P in terms of $ \hat{i} $ and $ \hat{j} $

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Answer

To find the displacement from P to R, we can again use the equations of motion:
[ \mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2 ]
Substituting the values, we have
[ \mathbf{s} = \left(24\hat{i} + 74\hat{j}\right)(4) + \frac{1}{2}(0\hat{i} - 10\hat{j})(4^2) ]
Calculating this gives us
[ \mathbf{s} = 96\hat{i} + 296\hat{j} ]

Step 4

(iv) the value of k, given that the particle reaches S after 16 seconds of motion.

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Answer

Using the equations of motion again, we find:
[ -k = 70(16) - 5(16)^2 ]
Solving this results in:
[ -k = 1120 - 1280 ]
Thus,
[ k = 160 ]

A particle is projected from a point P, as shown in the diagram, with an initial speed of 74 m s⁻¹ at an angle β to the horizontal, where \( \tan \beta = \frac{1}{2} \) - Leaving Cert Applied Maths - Question 3 - 2016

Worked Solution & Example Answer:A particle is projected from a point P, as shown in the diagram, with an initial speed of 74 m s⁻¹ at an angle β to the horizontal, where \( \tan \beta = \frac{1}{2} \) - Leaving Cert Applied Maths - Question 3 - 2016

(i) the initial velocity of the particle in terms of \( \hat{i} \) and \( \hat{j} \)

(ii) the velocity of the particle at point Q in terms of \( \hat{i} \) and \( \hat{j} \)

(iii) the displacement of R from P in terms of \( \hat{i} \) and \( \hat{j} \)

(iv) the value of k, given that the particle reaches S after 16 seconds of motion.

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