## Statement

A stone is let to free fall to the bottom of a cliff with a height of 80 m. A second later a second stone is thrown downward so that it reaches the bottom at the same time as the first one

- What was the launch velocity of the second stone?
- What was the velocity of the first stone when they both hit the bottom?
- How long was the second stone in the air?

## Solution

## Preliminary considerations

We are faced with a problem that combines a free fall motion (first stone) with a downward vertical launch (second stone). Both are uniformly accelerated rectilinear motions (u.a.r.m.) taking place in the vertical axis (*y-axis*). So, we will only have to use the following two equations:

We place the *system of reference* at the bottom of the cliff, as shown in the following figure:

The following table shows the *data* given in the statement:

Stone I | Stone II | |
---|---|---|

Final velocity v |
? | ? |

Initial velocity v_{0} |
0 | ? |

Accelerationn a |
g = -10 m/s^{2} |
g = -10 m/s^{2} |

Time t |
t_{1} |
t_{2 }= t_{1} - 1 |

Final position y |
0 | 0 |

Initial positiony_{0} |
80 | 80 |

We start by substituting the data in the equation of position of the u.a.r.m. for the first stone and we get the time that it takes the first stone to reach the ground.

$$0=80+\frac{1}{2}(-10){{t}_{1}}^{2}\Rightarrow {t}_{1}=\sqrt{\frac{80\xb72}{10}}=4\text{sg}$$

For the calculation of the velocity, we simply apply the equation of velocity of the u.a.r.m. using the data in the table, and considering the value of *t _{1}* = 4 s:

$${v}_{f1}=(-10)\xb74=\hspace{0.17em}-40\text{m/s}$$

The negative velocity indicates the downward direction of the stone. With this, we have responded the second question.

Since the second stone arrives to the ground same time as the first one but is released a second later, the time that it is in the air is a second less that the first one, therefore:

$${t}_{2}={t}_{1}-1=\hspace{0.17em}4-1=3\text{sg}$$

With this, we have responded the third question.

To answer the first question, we simply apply the equation of position in the u.a.r.m. for the second stone, considering the value of *t _{2}* = 3 sg:

$$0=80+{v}_{02}\xb73+\frac{1}{2}(-10)(3{)}^{2}\Rightarrow {v}_{02}=\frac{45-80}{3}=-11.6\text{m/s}$$

With this we answer the first question.