Answer
$t_{in\ air}\approx17.35\mbox{s}$
Work Step by Step
Using equation $H_{freely}=\frac{1}{2}gt_{freely}^2$ we can derive duration of falling freely.$$t_{freely}=\sqrt{\frac{2H_{freely}}{g}}$$
Now we can derive speed just before opening the parachute - maximum speed during given flight.$$v_{max}=gt_{freely}=\sqrt{2gH_{freely}}$$
The next step is to find the difference between maximal and final speed.$$\Delta v=v_{max}-v_{final}=\sqrt{2gH_{falling}}-v_{final}$$
Duration of flight using parachute can be derived using equation $\Delta v=a_{decel}t_{decel}$
$$t_{decel}=\frac{\Delta v}{a_{decel}}=\frac{\sqrt{2gH_{falling}}-v_{final}}{a_{decel}}$$
Knowing that duration of full flight is the duration of both free fall and decelerating, we can derive time in air.
$$t_{in\ air}=t_{freely}+t_{decel}=\sqrt{\frac{2H_{freely}}{g}}+\frac{\sqrt{2gH_{falling}}-v_{final}}{a_{decel}}=\sqrt{\frac{2\times50\mbox{m}}{9.8\frac{\mbox{m}}{\mbox{s}^2}}}+\frac{\sqrt{2\times{9.8}\frac{\mbox{m}}{\mbox{s}^2}\times50\mbox{m}}-3\frac{\mbox{m}}{\mbox{s}}}{2\frac{\mbox{m}}{\mbox{s}^2}}\approx17.35\mbox{s}$$
