A particle is projected horizontally from a point O at a height of 45 m vertically above a point P on level ground. It hits the ground at a point Q with PQ = 15 m. Take $g = 9.81$ ms$^{-2}$.

The particle is projected horizontally, so its initial vertical velocity is zero and the vertical motion is a free fall through 45 m.

Using $s = ut + \tfrac{1}{2}at^{2}$ with $u = 0$, $a = 9.81$ and $s = 45$,

$$45 = \tfrac{1}{2}(9.81)t^{2} \quad \Rightarrow \quad t^{2} = \frac{90}{9.81} = 9.1743 \quad \Rightarrow \quad t = 3.03\ \text{s}.$$

The horizontal distance of 15 m plays no part here: the two components of the motion are independent, and only the vertical one decides when the particle lands.