49. A waterworks' reservoir contains 400 tons of water. The waterworks can inject 60 tons of water into the reservoir per hour, while the reservoir continuously supplies water to residential areas, with the total water supply in $t$ hours being $120 \sqrt{6 t}$ tons $(0 \leqslant t \leqslant 24)$.
(1) From the start of water supply to which hour, is the water storage in the reservoir the least? What is the minimum amount of water in tons?
(2) If the water storage in the reservoir is less than 80 tons, a water shortage will occur. How many hours in a 24-hour day will a water shortage occur?
49. (1) Let the amount of water in the reservoir after $t$ hours be $y$ tons, then $y=400+60 t-120 \sqrt{6 t}$;

Let $\sqrt{6 t}=x$; then $x^{2}=6 t$, i.e., $y=400+10 x^{2}-120 x=10(x-6)^{2}+40$, so when $x=6$, i.e., $t=6$, $y_{\text {min }}=40$, which means from the start of water supply to the 6th hour, the water level in the reservoir is the lowest, only 40 tons.
(2) According to the problem, $400+10 x^{2}-120 x<80$, solving this gives $4<x<8$, i.e., $4<\sqrt{6 t}<8, \frac{8}{3}<t<\frac{32}{3}$, thus $\frac{32}{3}-\frac{8}{3}=8$, so there are approximately 8 hours of water shortage each day.