Presumably, the ball is kicked parallel to the ground below the cliff, so its altitude y at time t is
[tex]y(t)=50\,\mathrm m-\dfrac12gt^2[/tex]
where g = 9.80 m/s^2 is the acceleration due to gravity.
The ball hits the ground when y = 0:
[tex]0 = 50\,\mathrm m-\dfrac12gt^2[/tex]
[tex]t^2=\dfrac{100\,\mathrm m}g[/tex]
[tex]t=\dfrac{10}{9.80}\,\mathrm s\approx\boxed{3.2\,\mathrm s}[/tex]
