As we know that it is the condition of free fall when we remove the plank below the student
So here we know that
[tex]\Delta y = v_i t + \frac{1}{2}at^2[/tex]
here we have
[tex]v_i = 0[/tex]
[tex]a = 9.8 m/s^2[/tex]
now we have
[tex]8 = 0 + \frac{1}{2}(9.8)t^2[/tex]
[tex]t^2 = 1.63 [/tex]
[tex]t = 1.28 s[/tex]
so it will take 1.28 s to hit the water