## Essential University Physics: Volume 1 (4th Edition) Clone

We know that $t=\sqrt{\frac{2h}{g}}$ $t=\sqrt{\frac{2(1.6-0.93)}{9.8}}=0.3698s$ Now we can find the horizontal speed of the water as $v=\frac{x}{t}$ We plug in the known values to obtain: $v=\frac{2.1}{0.3698}=5.7\frac{m}{s}$