17. Una piedra se deja caer libremente al fondo de un precipicio de 80 m de altura. Un segundo más tarde, una segunda piedra se lanza hacia abajo de tal forma que alcanza la primera justamente cuando ésta llega al fondo.

a. ¿Con qué velocidad se lanzó la segunda piedra?
b. ¿Qué velocidad llevaba la primera piedra cuando fue alcanzada?
c. ¿Cuánto tiempo dura en el aire la segunda piedra?

The answer involves calculating velocities and times for rocks dropped from a cliff. The initial and reached velocities are the same, and the time for the second rock in the air is determined.

Explanation:

a. To determine the speed at which the second rock was thrown, we can use the kinematic equation:
v = u + gt, where u is the initial velocity (0 m/s), g is the acceleration due to gravity (-9.81 m/s^2), and t is the time taken to reach the bottom (3.5 seconds). This gives us a velocity of 31.035 m/s.

b. When the first rock is reached by the second one, the velocity of the first rock is equal to the velocity of the second. Therefore, the velocity of the first rock when it was reached is 31.035 m/s.

c. For the time the second rock is in the air, we can consider either the time it takes to hit the bottom (3.5 seconds) or the total time both rocks are falling (4.5 seconds). So, the time for the second rock in the air is 3.5 seconds.

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