Se deja caer una piedra A en reposo desde un acantilado muy alto. Cuando ha caído 5 m, se deja caer una piedra B. A. Explicar ¿cómo cambia la distancia entre ambas piedras a medida que caen? B. Determina la velocidad de la piedra B cuando ha recorrido 5 m. C. En una sola gráfica velocidad vs tiempo, bosque las gráficas de las dos piedras cuando la piedra B ha recorrido 10 m. NO necesita colocar valores en las gráficas.

Here's what I get  

Explanation:

A. Distance between A and B.

h = -½gt²

The stones go faster the farther they fall.

Stone A has already reached 5 m when B is released.

When B reaches 5 m, A has dropped further and is falling even faster.

The distance between the stones increases with time.

Figure 1 shows this effect in a graph of height vs. time.

B. Speed of Stone B

v² = 2gh =2 × ( -9.81 m·s⁻²) × (-5 m) = 98.1 m·s⁻²  

v = 9.9 m/s

The stone is travelling at 9.9 m/s when it reaches 5 m.

C. Velocity vs time

v = -gt

Both stones accelerate at the same rate.

When Stone B has reached 10 m at time t, Stone A is falling much faster.

Fig. 2 shows this in a graph of velocity vs time.

the answer for apex is the lowest