At what distance from the target should the helicopter drop the cargo if it flies at an altitude of 80 m with a horizontal speed of 108 km / h? What is the speed of the load when it hits the target?
h = 80 m.
Vх = 108 km / h = 30 m / s.
g = 9.8 m / s ^ 2.
The load, when falling, moves horizontally uniformly at a speed Vx, vertically with an acceleration of gravity g.
The time of the fall of the load t is found by the formula: t = √ (2 * h / g).
t = √ (2 * 80 m / 9.8 m / s ^ 2) = 4 s.
Vу = g * t.
Vу = 9.8 m / s ^ 2 * 4 s = 39.2 m / s.
The impact speed of the load at the moment of falling is expressed by the formula: V = √ (Vу ^ 2 + Vх ^ 2).
V = √ ((39.2 m / s) ^ 2 + (30 m / s) ^ 2) = 49.4 m / s.
The distance to the target is expressed by the formula: L = √ (h ^ 2 + Lx ^ 2).
Lx = Vx * t.
Lx = 30 m / s * 4 s = 120 m.
L = √ ((80m) ^ 2 + (120m) ^ 2) = 144m.
Answer: L = 144 m, V = 49.4 m / s.
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