A stone is thrown horizontally from a building of height 45 m with an initial speed of 20 m/s. How far from the foot of the building does it land?

The initial vertical velocity ($u_y$) is $0 \text{ m/s}$ because the stone is thrown horizontally. The vertical displacement ($s_y$) is the height of the building, $45 \text{ m}$. The acceleration due to gravity ($a_y$) is $9.8 \text{ m/s}^2$.
We use the kinematic equation: $s_y = u_y t + \frac{1}{2}a_y t^2$.

$$45 \text{ m} = (0 \text{ m/s})t + \frac{1}{2}(9.8 \text{ m/s}^2)t^2$$ $$45 = 4.9t^2$$ $$t^2 = \frac{45}{4.9}$$ $$t = \sqrt{\frac{45}{4.9}}$$ $$t \approx 3.03 \text{ s}$$

The horizontal velocity ($v_x$) is constant at $20 \text{ m/s}$. The horizontal distance ($x$) is given by $x = v_x t$.

$$x = (20 \text{ m/s})(3.0304578 \text{ s})$$ $$x \approx 60.6 \text{ m}$$

The stone lands approximately $60.6 \text{ m}$ from the foot of the building.