A load was dropped from a helicopter flying horizontally at an altitude of 320m at a speed of 50m / s.

A load was dropped from a helicopter flying horizontally at an altitude of 320m at a speed of 50m / s. a) how long does the load awaken to fall (air resistance is negligible)? b) how far will the load fly horizontally during the fall? c) at what speed the load will fall to the ground

Data: h (helicopter height, load path) = 320 m; Vg (helicopter speed, horizontal cargo speed) = 50 m / s; Vw1 (vertical speed of the load before dropping) = 0 m / s.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) Drop time: h = Vв1 * t + a * t ^ 2/2 = 0 + g * t ^ 2/2 and t = √ (2h / g) = √ (2 * 320/10) = 8 s …

2) Distance from the point of discharge: S = Vg * t = 50 * 8 = 400 m.

3) Velocity at the moment of falling: V = √ (Vv2 ^ 2 + Vg ^ 2) = √ ((g * t) ^ 2 + Vg ^ 2) = √ ((10 * 8) ^ 2 + 50 ^ 2) = 94.34 m / s.