A particle is projected horizontally at 10ms from the height of 45m, calculate the horizontal distance covered by the particle before hitting the ground. (g = 10ms−2)

For vertical motion, the initial vertical velocity is $u_y = 0 \, \text{m/s}$. The vertical displacement is $h = 45 \, \text{m}$, and the acceleration due to gravity is $g = 10 \, \text{m/s}^2$.
Using the kinematic equation $s = ut + \frac{1}{2}at^2$:

$$h = u_y t + \frac{1}{2}gt^2$$ $$45 \, \text{m} = (0 \, \text{m/s})t + \frac{1}{2}(10 \, \text{m/s}^2)t^2$$ $$45 = 5t^2$$ $$t^2 = \frac{45}{5}$$ $$t^2 = 9 \, \text{s}^2$$ $$t = \sqrt{9 \, \text{s}^2}$$ $$t = 3 \, \text{s}$$

The horizontal velocity $u_x = 10 \, \text{m/s}$ remains constant throughout the flight. The horizontal distance $x$ is given by:

$$x = u_x t$$ $$x = (10 \, \text{m/s})(3 \, \text{s})$$ $$x = 30 \, \text{m}$$

The horizontal distance covered by the particle is $\text{30 m}$.