Question

Old Faithful geyser in Yellowstone National Park shoots water every hour to a height of 40.0 m. With what velocity does the water leave the ground? (Assume no air resistance and that g 9.81 m/s2)

Answer 1

Vo = 28 [m/s]

Step-by-step explanation:

To solve this problem, use the following equation of kinematics.

`v_(f)^(2)= v_(o)^(2)-(2*g*y)`

where:

Vf = final velocity = 0

Vo = initial velocity [m/s]

g = gravity acceleration = 9.81 [m/s²]

y = elevation = 40 [m]

Note: when the geyser water reaches the maximum height, the speed is zero. Another important observation is that the negative sign in the equation indicates that the movement of water is contrary to the direction of the acceleration of gravity.

0 = Vo² - (2*9.81*40)

Vo = (784.8)⁰⁵

Vo = 28 [m/s]