H = 19.6 meters – the height to which the water jet rises;
g = 9.8 m / s – acceleration of gravity.
It is required to determine v (m / s) – the speed at which the water is ejected by the pump.
Let some part of the water of mass m be ejected by the pump. At this moment, it has kinetic energy Wkin, thanks to which it rises to the height H. Then, according to the law of conservation of energy:
Wkin = Wpot;
m * v ^ 2/2 = m * g * h;
v ^ 2/2 = g * h;
v = (2 * g * h) ^ 0.5 = (2 * 9.8 * 19.6) ^ 0.5 = 384.2 ^ 0.5 = 19.6 m / s.
Answer: water from the pump is thrown out at a speed of 19.6 m / s.
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