[tex]d=\dfrac{1}{2}at^2\\\\We\ have:\\\\d=50\ ft,\ a=3.2\dfrac{ft}{s^2}\\\\Substitute:\\\\\dfrac{1}{2}\left(3.2\dfrac{ft}{s^2}\right)t^2=50\ ft\\\\1.6\dfrac{ft}{s^2}t=50\ ft\qquad\text{divide both sides by}\ 1.6\dfrac{ft}{s^2}\\\\t^2=50\ ft\cdot\dfrac{1}{1.6}\dfrac{s^2}{ft}\\\\t^2=\dfrac{50}{1.6}s^2\\\\t^2=\dfrac{500}{16}s^2\to t=\sqrt{\dfrac{500}{16}s^2}\\\\t=\dfrac{\sqrt{500}}{\sqrt{16}}s\\\\t=\dfrac{\sqrt{100\cdot5}}{4}s\\\\t=\dfrac{\sqrt{100}\cdot\sqrt5}{4}s\\\\t=\dfrac{10\sqrt5}{4}s\approx5.59\ s[/tex]
[tex]Answer:\ \boxed{5.59\ s}[/tex]