Answer to Question #94522 in Mechanics | Relativity for Kate

Question

Suppose you adjust your garden hose for a faster stream of water. You point the nozzle vertically up at a height of 1.8 m above the ground. When you quickly turn off the nozzle, you hear the water striking the ground next to you for 2.5 seconds. What is the speed of the water as it leaves the nozzle?

Answer 1

$y = y_0 + v_0t - \frac{gt^2}{2} = 0; \\ v_0 = \frac{gt}{2} - \frac{y_0}{t} = \frac{9.8\frac{\text{m}}{\text{s}^2}\cdot2.5\text{ s}}{2} - \frac{1.8\text{ m}}{2.5\text{ s}} = 11.53\frac{\text{m}}{\text{s}}.$