A stone is thrown horizontally from the top of a building of height h with an initial velocity of 10 m/s. It hits the ground at a horizontal distance of 50 m from the building. Calculate the height of

The horizontal motion of the stone is uniform, meaning its horizontal velocity remains constant.
We use the formula:

$$\text{Horizontal distance} = \text{Horizontal velocity} \times \text{Time}$$

Given:
Horizontal distance ($R$) = $50 \text{ m}$
Horizontal velocity ($u_x$) = $10 \text{ m/s}$

$$R = u_x t$$ $$50 \text{ m} = (10 \text{ m/s}) t$$ $$t = \frac{50 \text{ m}}{10 \text{ m/s}}$$ $$t = 5 \text{ s}$$

The vertical motion of the stone is under constant acceleration due to gravity. The initial vertical velocity is $0 \text{ m/s}$ because the stone is thrown horizontally.
We use the kinematic equation:

$$h = u_y t + \frac{1}{2} g t^2$$

Given:
Initial vertical velocity ($u_y$) = $0 \text{ m/s}$
Acceleration due to gravity ($g$) = $10 \text{ m/s}^2$ (as indicated in the student's work)
Time ($t$) = $5 \text{ s}$

$$h = (0 \text{ m/s})(5 \text{ s}) + \frac{1}{2} (10 \text{ m/s}^2) (5 \text{ s})^2$$ $$h = 0 + \frac{1}{2} (10 \text{ m/s}^2) (25 \text{ s}^2)$$ $$h = (5 \text{ m/s}^2) (25 \text{ s}^2)$$ $$h = 125 \text{ m}$$

The height of the building is $\text{125 m}$.