1. ## Help with physics problem please?

Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5m above the ground. When you quickly move away from the vertical you hear the water striking the ground next to you for another 2.0s. What is the water speed as it leaves the nozzle?

How should I approach this problem? How do I know what equations I should be using for this type of problem?

2. I suppose that you are opening the tap when the nozzle is already placed vertically above the ground.

Let's take the upwards position as the positive direction.
The initial speed of the water is positive,
The acceleration due to gravity is negative,
The displacement of the water is negative (ie, final position is lower than initial position)

Those settled, you can use:

$\displaystyle s = ut + \frac12 at^2$

Where s is the displacement,
u the initial velocity of the water,
t the time taken to cover that displacement,
a the acceleration the water undergoes.

So, we get:

$\displaystyle -1.5 = u(2) + \frac12 (-9.8)(2)^2$

Solve for u.

I hope it helps!

EDIT:

Concerning the equations, there are a couple that you need to know:

$\displaystyle v = u+at$

$\displaystyle s = ut + \frac12 at^2$

$\displaystyle v^2 = u^2 + 2as$