The helicopter takes off from the airfield vertically with an acceleration of 3m / s2. After a while, the pilot turned

| May 7, 2021 | education

The helicopter takes off from the airfield vertically with an acceleration of 3m / s2. After a while, the pilot turned off the engine. The sound on the ground at the take-off site ceased to be heard after a time of 30 s. What was the speed of the helicopter when the engine was turned off? Take the speed of sound 320 m / s.

The distance the sound travels will be equal to the altitude the helicopter will climb.

The path that the sound has traveled:

S1 = V1 * t, where V1 is the speed of sound (V1 = 320 m / s), t is the time of hearing the sound (t = 30 s).

The altitude to which the helicopter climbed:

S = (V ^ 2 – V0 ^ 2) / 2a, where V is the speed of the helicopter when the engine is turned off, V0 is the initial speed of the helicopter (V0 = 0 m / s), and is the acceleration of the helicopter (a = 3 m / s2) …

V1 * t = V ^ 2 / 2a.

V = √ (V1 * t * 2а) = √ (320 * 30 * 2 * 3) = 240 m / s.



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