## Statement

A glass of water on the edge of a table falls towards the floor from a height of 1.5 m. Considering that gravity is 10 m/s^{2}, calculate:

a) The time the glass is in the air.

b) The velocity with which it impacts on the ground.

## Solution

**Question a)**

**Data**

H = 1.5 m

When it gets to the ground y = 0 m.

g = 10 m/s^{2}

**Resolution**

To answer this question simply apply the position equation in free fall and solve for time when the glass is at y = 0 m, that is, when it reaches the ground:

$$y=H-\frac{g\xb7{t}^{2}}{2}\Rightarrow \phantom{\rule{0ex}{0ex}}t=\sqrt{\frac{-2\xb7(y-H)}{g}}\Rightarrow \phantom{\rule{0ex}{0ex}}t=\sqrt{\frac{-2\xb7(0-1.5)}{10}}\Rightarrow \phantom{\rule{0ex}{0ex}}t=\sqrt{\frac{3}{10}}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)t=0.55}$$

**Question b)**

**Data**

H = 1.5 m

When it gets to the ground y = 0 m.

g = 10 m/s^{2}

Time that it takes to reach the ground t = 0.55 s

**Resolution**

Since we know the time that it takes reaching the floor, it is enough to apply the equation of velocity for that time:

$$v=-g\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}v=-10m/{s}^{\overline{)2}}\xb70.55\overline{)s}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)v=5.5m/s}$$