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

answer and explanation;

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer