Verified answer

Of course let us neglect the force of friction due to air etc. Let us use the kinematic equation, the third one ie v^2 = u^2 --2 gs

Here v = 0 at max height, g = 9.8 m/s^2, s = 40m and u =?

Plugging we get, u^2 = 196*4

There fore u = sq rt (4*196) = 2*14 = 28 m/s

Velocity with which water leave the ground is 28m/s.

While answering your question, I am neglecting two factors. The first one is air resistance and the second one is reduction in gravitational pull at 40.0 meter height. At 40 meter height gravitational pull is smaller by very small amount that is g1 = g r^2 / r1^2 where g1 is gravitational pull at 40 meter height and r is radius of the earth and r1 = r+40. Neglecting above two small correction factors, the answer is as follow.

The geyser is ejecting the water with velocity V, hence its kinetic energy is mV2/2 when water droplets reach at the height of 40 meter its potential energy is mgh where g = 9.81 m/sec^2

These both energies are equal hence can be equated as follow.

(mV^2) / 2 = mgh

V^2 = gh

V = Sq. Root 2(gh)

V = Sq. Root 2(9.81 x 40)

V = 28.01 m/sec

The water is shooting up at 28.01 m/sec

You can add some correction factor for air resistance.