A helicopter takes off along the vertical with an acceleration `a = 3m//s^(2)` and zero initial velocity. In a certain time the pilot switches off the engine. At the point of take off, the sound dies away in a time `t_(2) = 30 sec`. Determine the velocity of the helicopter at the moment when its engine is switched off assuming that velocity of sound is 320 m/s.

Answer

`v = 80 m//sec`

Explanation

Distance moved by helicopter before its engine is stopped `= (1)/(2) a t_(1)^(2) = (3)/(2) t_(1)^(2)`
Time taken by sound to reach ground `= ((3)/(2) t_(1)^(2))/(320)`
So, `((3)/(2) t_(1)^(2))/(320) + t_(1) = 30`
`3 t_(1)^(2) + 640 t_(1) - 19200 = 0`
`t_(1) = (80)/(3) s`
So, speed of helicopter when engine is stopped,
`v = at_(1) = 3 xx (80)/(3) = 80 m//s`