t = 3 seconds – the time after which the arrow reached the maximum point of the trajectory;
H = 20 meters – maximum boom flight height;
g = 10 m / s ^ 2 – acceleration of gravity.
It is required to determine v0 (m / s) – the initial speed of the arrow.
According to the law of conservation of energy, the kinetic energy of the arrow at the beginning of the flight is equal to the potential energy of the arrow at the maximum point of the trajectory:
Ekinetic = Epotential;
m * v0 ^ 2/2 = m * g * H;
v0 ^ 2/2 = g * H;
v0 ^ 2 = 2 * g * H;
v0 = (2 * g * H) ^ 0.5 = (2 * 10 * 20) ^ 0.5 = (20 * 20) ^ 0.5 = 20 m / s.
Answer: the initial speed of the boom is 20 m / s.
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