A particle is projected from point P, as shown in the diagram, with initial speed 82 m s⁻¹ at an angle of $ heta = an^{-1}(rac{40}{9})$ to the horizontal, where tan $ heta = rac{40}{9}$. P is a point on the horizontal roof of a tall building of height h. Q is the highest point reached by the particle. The particle lands at R, a point on the horizontal roof of another building of height 30 m, as shown. Find (i) the initial velocity of the particle in terms of i and j (ii) the velocity of the particle after 5 seconds of motion in terms of i and j (iii) the displacement of Q from P in terms of i and j (iv) the value of h, given that the particle lands at R after 17 seconds of motion. The street between the buildings is horizontal and is 54 m wide. (v) Find the time for which the particle is not passing over a building.

Leaving Cert Applied Maths Projectiles: A particle is projected from poi

A particle is projected from point P, as shown in the diagram, with initial speed 82 m s⁻¹ at an angle of $ heta = an^{-1}(rac{40}{9})$ to the horizontal, where tan $ heta = rac{40}{9}$ - Leaving Cert Applied Maths - Question 3 - 2019

Question 3

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A particle is projected from point P, as shown in the diagram, with initial speed 82 m s⁻¹ at an angle of $ heta = an^{-1}(rac{40}{9})$ to the horizontal, where ta... show full transcript

Worked Solution & Example Answer:A particle is projected from point P, as shown in the diagram, with initial speed 82 m s⁻¹ at an angle of $ heta = an^{-1}(rac{40}{9})$ to the horizontal, where tan $ heta = rac{40}{9}$ - Leaving Cert Applied Maths - Question 3 - 2019

Step 1

Find (i) the initial velocity of the particle in terms of i and j

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Answer

The initial velocity can be broken down into its horizontal and vertical components using trigonometric functions:

Horizontal component: $u_x = u imes ext{cos}( heta) = 82 imes rac{9}{ ext{sqrt}(9^2 + 40^2)}$
Vertical component: $u_y = u imes ext{sin}( heta) = 82 imes rac{40}{ ext{sqrt}(9^2 + 40^2)}$

Hence, the initial velocity in vector form is:

$extbf{u} = u_x extbf{i} + u_y extbf{j}$

Where, $extbf{u} = 82 ext{cos}( heta) extbf{i} + 82 ext{sin}( heta) extbf{j}$ .

Step 2

Find (ii) the velocity of the particle after 5 seconds of motion in terms of i and j

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Answer

To find the velocity after 5 seconds, we use the components found previously:

The horizontal component remains constant: $v_x = u_x = 82 imes rac{9}{ ext{sqrt}(9^2 + 40^2)}$
The vertical component changes due to gravity: $v_y = u_y - gt = u_y - 9.8 imes 5$

Thus, the velocity vector at t = 5 seconds is:

$extbf{v} = v_x extbf{i} + v_y extbf{j}$

Step 3

Find (iii) the displacement of Q from P in terms of i and j

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Answer

The displacement can be found by considering the horizontal distance traveled and the vertical height of point Q.

Horizontal distance $(d_x)$ traveled in 17 seconds: $d_x = u_x imes t = 82 imes rac{9}{ ext{sqrt}(9^2 + 40^2)} imes 17$
Vertical distance $(d_y)$ will equal the height difference: $d_y = h$ .

Thus, the displacement in vector form is:

$ext{Displacement} = d_x extbf{i} + d_y extbf{j}$

Step 4

Find (iv) the value of h, given that the particle lands at R after 17 seconds of motion.

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Answer

At t = 17 seconds, we set up equations for the horizontal and vertical motions:

Horizontal motion: $54 = u_x imes 17$
Vertical motion: $h - 30 = u_y imes t - rac{1}{2} g t^2$

Solving both will give us the value of h.

Step 5

Find (v) the time for which the particle is not passing over a building.

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Answer

We analyze the vertical motion:

The particle reaches its peak height at $t_{ ext{peak}} = rac{u_y}{g}$ .
Then it starts descending.
To find the time it remains above the height of the building:
- Set the height equal to the height of the building and solve for time.

Thus, we can determine the intervals of time when the particle is above the building.

A particle is projected from point P, as shown in the diagram, with initial speed 82 m s⁻¹ at an angle of $ heta = an^{-1}( rac{40}{9})$ to the horizontal, where tan $ heta = rac{40}{9}$ - Leaving Cert Applied Maths - Question 3 - 2019

Worked Solution & Example Answer:A particle is projected from point P, as shown in the diagram, with initial speed 82 m s⁻¹ at an angle of $ heta = an^{-1}( rac{40}{9})$ to the horizontal, where tan $ heta = rac{40}{9}$ - Leaving Cert Applied Maths - Question 3 - 2019

Find (i) the initial velocity of the particle in terms of i and j

Find (ii) the velocity of the particle after 5 seconds of motion in terms of i and j

Find (iii) the displacement of Q from P in terms of i and j

Find (iv) the value of h, given that the particle lands at R after 17 seconds of motion.

Find (v) the time for which the particle is not passing over a building.

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A particle is projected from point P, as shown in the diagram, with initial speed 82 m s⁻¹ at an angle of $ heta = an^{-1}(rac{40}{9})$ to the horizontal, where tan $ heta = rac{40}{9}$ - Leaving Cert Applied Maths - Question 3 - 2019

Worked Solution & Example Answer:A particle is projected from point P, as shown in the diagram, with initial speed 82 m s⁻¹ at an angle of $ heta = an^{-1}(rac{40}{9})$ to the horizontal, where tan $ heta = rac{40}{9}$ - Leaving Cert Applied Maths - Question 3 - 2019