## Statement

## Solution

**Data**

g = 9.8 m/s^{2}

t = 8 s

**Resolution**

By applying the formula of velocity in free fall, we get that:

$$v=-g\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}v=-9.8m/{s}^{\overline{)2}}\xb78\overline{)s}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)v=-78.4m/s}$$

To find out the distance traveled, we must calculate the displacement of the stone (Δy). This displacement is calculated by subtracting the position after 8 second (y8) from the position at the time 0 (y0). If we develop the equations of the position in both cases:

$$\Delta y={y}_{8}-{y}_{0}=H-\frac{1}{2}\xb7g\xb7{{t}_{8}}^{2}-\left(H-\frac{1}{2}\xb7g\xb7{{t}_{0}}^{2}\right)\Rightarrow \phantom{\rule{0ex}{0ex}}\Delta y=H-\frac{1}{2}\xb7g\xb7{8}^{2}-\left(H-\overline{)\frac{1}{2}\xb7g\xb7{8}^{2}}\right)\Rightarrow \phantom{\rule{0ex}{0ex}}\Delta y=\overline{)H}-\frac{1}{2}\xb79.8\xb7{8}^{2}-\overline{)H\Rightarrow}\phantom{\rule{0ex}{0ex}}\overline{)\Delta y=313.6m}$$