Question

The Steamboat aloftser in Yellowstone National Park, Wyoming is capable of shooting its hot water up from the ground with a speed of 48.0 m/s. How high can this geyser shoot? (HINT: USE KINEMATIC EQUATIONS)

Answer 1

use energy conservation

kinetic energy is converted into potential here.

Ekin = 1/2mv^2
Epot =m*g*h
1/2V^2 = g*h

Answer 2

Im confused, can you explain it more?

Answer 3

You conserve energy here..... the kinetic energy in the start is

1/2*m*v^2

ALL this energy is converted to potential energy. The potential energy at the end is given as:

m*g*h

You know they are equal, so we can use an equation to calculate the height h.

m*g*h = m*1/2*V^2

We see that we can take m out:

g*h = 1/2 V^2

Answer 4

So you cant do anything other than that?

Answer 5

Yes, you can... but this is the easiest way to do it.

You can also solve by using

s = 1/2*a*t^2..

a = g = 9.81 m/s2
t = v/g = 48/9.81

s = 1/2 * a * (v/g)^2 = 9.81*(48/9.81)^2

Answer 6

But it boils down into the same equation as:

s = h = 1/2 * g * (v/g)^2 = 1/2 *V^2/g

Answer 7

Okay, thank you ^^