43. The Old Faithful Geyser shoots water into the air every hour to a height 3 points of 50m. Calculate the velocity at which the water leaves the ground. V=V[ 2gh g = 9.8 m/s ^ 2 your answer to one decimal place.

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.

One answer could be erosion

At maximum height, kinetic energy is converted in potential energy  formula: at max height, ke = ½ * 0.27kg * (16m/s) ² = 34.56 j  so the remaining pe = (41.70 – 34.56)= 7.14 j = 0.27 kg * 9.8m/s² * h  = (9.8/7.14 * 0.27)so max height h = 5.08 or 5 m 

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