I mean for the distance from the former question ☝️

For horizontal motion, the displacement $S$ becomes the horizontal distance $R$, the initial velocity $u$ becomes the initial horizontal velocity $u_x$, and the acceleration $a$ becomes the horizontal acceleration $a_x$.

$$R = u_x t + \frac{1}{2}a_x t^2$$

In projectile motion, assuming no air resistance, the horizontal acceleration $a_x$ is $0 \, \text{m/s}^2$.

$$R = u_x t + \frac{1}{2}(0)t^2$$ $$R = u_x t$$

From the previous problem:
Initial horizontal velocity $u_x = 10 \, \text{m/s}$
Time taken to reach the ground $t = \sqrt{10} \, \text{s}$ (calculated in part a)

$$R = (10 \, \text{m/s})(\sqrt{10} \, \text{s})$$ $$R = 10\sqrt{10} \, \text{m}$$ $$R \approx 31.62 \, \text{m}$$

The horizontal distance (c) is $\text{31.6 m}$.