Water shot from a hose reaches 2.5 m height

[Jacob87411]

A Hose shoots water straight up for a distance of 2.5m. The end opening on the hose has an area of .75cm^2. (A) What is the speed of the water as it leaves the hose? (B) How much water comes out in 1 minute.

If I have A i can find B, I just can't figure out how to find A. I know gravity is going to be what is pulling it down, I just am unsure of what equation to use. I was thinking maybe Bernouli's Equation but not positive, any help is appreciated

 

[Doc Al]

Treat the water as any other "projectile". What speed does a projectile need to rise up that distance?

 

[Jacob87411]

Sorry for lack of symbols, still figuring it out how to work them all but..

Y = Y0 + V0T + -1/2GT^2

Y0 = Original Height
V0 = Original Velocity

So..

2.5 = V0 - .5(9.8)

So 2.5 + 4.9 which is 7.4?

I may of mixed up the equation, forgot book in my locker which is part of the problem

 

[Doc Al]

Y = Y0 + V0T + -1/2GT^2

Y0 = Original Height
V0 = Original Velocity

That equation relates distance and time. But you aren't given the time.

You need a kinematic equation that relates distance and speed.

 

[Jacob87411]

V^2 = V0^2 - 2aY or
V^2 = V0^2 + 2a(Y1 - Y0)

The water at its very peak V will equal 0 won't it? So..

0 = V0^2 - 2(9.8)(2.5 - 0)
0 = V0^2 - 49
V0^2 = 49
V0 = 7

 

[Doc Al]

Right! But don't forget units: 7 m/s.

 

[Jacob87411]

Thanks a lot, i really appreciate it

 

[pack_rat2]

A comment...I'd do that problem by considering a small "piece" of the fluid, finding its potential energy at its maximum height (m*g*h), and equating that to its initial kinetic energy.

 

[Doc Al]

An excellent approach which leads to the exact same equation:
$mgh = 1/2mv^2$ => $v^2 = 2gh$