Old Faithful geyser in Yellowstone Park shoots water hourly to a height of 30 m.With what velocity does the water leave the ground?

Question

Quizplus · Accepted Answer

We can use the equation for projectile motion: 
y = vi*t + (1/2)*a*t^2

Where y = 30 m (maximum height reached by the water), vi is the initial velocity, t is the time it takes for the water to reach its maximum height (0.0 seconds since the water starts at the ground), and a is the acceleration due to gravity (-9.8 m/s^2 or -g).

Thus, we can solve for vi:

30 m = vi*(0.0 s) + (1/2)*(-g)*(0.0 s)^2

vi = sqrt((2*g*y)/1)

Plugging in values, we get:

vi = sqrt((2*9.8 m/s^2 * 30 m)/1) 
vi = sqrt(588) 
vi = 24.2 m/s

Thus, to two significant figures, the water from Old Faithful geyser leaves the ground with a velocity of 24 m/s.

Old Faithful Geyser in Yellowstone Park Shoots Water Hourly to a Height