Assumption 1... the fish is near the surface.
So, the pelican has "fallen" 20 metres, under the force of gravity. The fish starts to move .10 of a second after seeing it, so the pelican needs to cover the next 5 metres in under 0.10 seconds.
A velocity vs time graph could show it, or maybe a displacement vs time graph of the last 5 metres? (with a negative gradient).
calculations
\(s_1=u_1t_1+\frac{at_1^2}{2}\), so \(-20=0+(\frac{-9.8}{2} )\times (t_1)^2\)
This means that \(t_1\), the time taken before the fish spots it, is equal to \(\sqrt{\frac {20}{4.9}} \quad \therefore \quad t=2.02\)
\(v_1=u_1+at_1\), so \(v_1= 0+9.8\times t_1\), meaning that \(v_1=19.7989899\) or \(19.80\)
\(v_1\) now becomes your \(u_2\)...
\((v_2)^2=(u_2)^2+2as_2\) so \(v=\sqrt {19.80^2 +19.6\times 5}\)
So \(v= ~7\sqrt {10}\)
Now, back into \(v=u+at\), we get that \(t= \frac {7\sqrt {10} - 19.80} {9.8} \quad \therefore \quad t=0.23s\), which is greater than \(0.1s\), so the Pelican goes hungry...
Jirachi got in before me, but I thought I had better keep mine up there...
One of us in wrong... 
Nvm. 0.23 same same
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