Cliff diving

$s=ut+\frac{1}{2}at^2~~~\Longrightarrow 50~\textrm{m}=0 \times t + \frac{1}{2} \times 9.8~\textrm{m/s}^2 \times t^2$

$\Longrightarrow t^2 \times 4.9~\textrm{m/s}^2 = 50~\textrm{m} ~~~\Longrightarrow t^2 = \frac{50~\textrm{m}}{4.9~\textrm{m/s}^2}$

$\therefore t = \sqrt{\frac{50~\textrm{m}}{4.9~\textrm{m/s}^2}}\approx 3.19~\textrm{seconds}$

apliy the motion secon equation and put u=0 and h=50 g=9.8 and find time

$S = S_o + V_o \cdot t + \frac{\alpha \cdot t^2}{2} \\\\ 50 = 0 + 0 + \frac{10t^2}{2} \\\\ 5t^2 = 50 \\\\ t^2 = 10 \\\\ \boxed{t = 3.19 \, \text{s}}$

since,S=ut+1/2at^2 , h=1/2gt^2(as,u=0)

We get,
            50=1/2*9.8*t^2
           50=4.9*t^2
          t^2=50/4.9=500/49
         rooting on both sides,we get,
        t=+or- 22.3/7
        And since time cannot be negative,
       Therefore,t=22.3/7
          So,t=3.1943828249997