Un niño lanza una piedra, desde un acantilado con una velocidad horizontal de 6 m/s. Calcule la distancia "x" y la velocidad con que se estrella la piedra, en el agua

    x = 6 √ 2y / 9.8 ,     v = √(2y g)

Explanation:

This is a projectile launching exercise.

There is no acceleration on the horizontal axis

          vₓ = x / t

           x = vₓ  t

on the axis and the initial speed is zero ( = 0)

       y = ½ g t²

       t = √ 2y / g

       t = √2 y / 9.8

this is the time to go down

      v = vo - gt

      v = 0 - g √ 2y / g

      v = √2y g

the horizontal distance is

        x = 6 √ 2y / 9.8

in order to complete the calculations you need at the height of the cliff

hi i’m sorry to call us today so we will get back with y’all and we will have to do it haha omg was the day

think its the third option