3. (a) A particle is projected horizontally with an initial speed of 14 m s⁻¹ from the top of a straight vertical cliff of height 80 m - Leaving Cert Applied Maths - Question 3 - 2015

After 4 seconds, the horizontal and vertical components of velocity can be found as follows:

1. Horizontal velocity (remains constant):

   $v_x = 14 m/s$

2. Vertical velocity (under gravity):

   $v_y = u + gt = 0 + 9.8 \times 4 = 39.2 m/s$

Thus, the velocity vector at point A:

$\vec{v}_A = 14 i + 39.2 j m/s$

The speed of the particle can be determined using the following formula:

$|\vec{v}| = \sqrt{(v_x^2 + v_y^2)}$

Substituting the values:

$|\vec{v}| = \sqrt{(14^2 + 39.2^2)} = \sqrt{196 + 1536.64} = \sqrt{1732.64} \approx 41.6 m/s$

To find the direction, we can use the tangent function:

$\beta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{39.2}{14}\right) \approx 70.5^{\circ} \text{ from the horizontal}$

To find the value of k, we need to analyze the displacement:

From point P to point B, the horizontal displacement is:

$ext{Displacement} = 189 i + k j$

Using the time of flight:

$k = 0 + \frac{1}{2}gt^2 = 0 + \frac{1}{2} \cdot 9.8 \cdot 4^2 = 78.4 m$

Since the total vertical distance from P is:

$s_k = k - 0 = k$

It follows that:

$k = 60 m - 5 \cdot 9^2 = 135$