A particle is projected from point P to point S, as shown in the diagram, with initial speed 136 m s⁻¹ at an angle of β° to the horizontal where tan β = \frac{15}{8} The particle passes vertically over the points Q and R during its motion - Leaving Cert Applied Maths - Question 3 - 2018

At maximum height, the vertical component of the velocity becomes zero. Thus, we set:

$v_y = \mathbf{u_y} - g t$ where (v_y = 0).$$

Solving for time, we can establish:

$0 = 120 - 10t$

Rearranging gives us:

$$10t = 120 \Rightarrow t = 12 s.$

The horizontal motion of the projectile covers:

$\mathbf{r} = u_x t$ (t) being the total time of flight.

The total time taken to reach the ground can be found, which is (t=24 s), so:

$$\text{Range} = 64 \cdot 24 = 1536 m.$

When the particle is above point Q, we will derive the height from the initial vertical displacement given by:

$\mathbf{y} = u_y t - \frac{1}{2}gt^2$ With |PQ| = 512 m and using the time to reach Q leading to:

$t = 8 s$

Substituting, we have:

$$\mathbf{y} = 120 \cdot 8 - \frac{1}{2} \cdot 10 \cdot 8^2 = 640 m.$

Using conservation of energy or direct component analysis at point R, we find:

$\mathbf{v} = \mathbf{u} - gt$ At (t = 14 s), we can find:

$\mathbf{v} = 64 \mathbf{i} + (120 - 10 \cdot 14)\mathbf{j}$ Thus, $\mathbf{v} = 64 \mathbf{i} - 20 \mathbf{j},$ For the speed magnitude,

$$|\mathbf{v}| = \sqrt{64^2 + (-20)^2} = 67.05 \text{ ms}^{-1}.$