A helicopter takes off along the vertical with an acceleration a = 3 $m / s^{2}$ and zero initial velocity. In a certain time $t_{1}$ , the pilot switches off the engine. At the point of take off sound dies away in 30 sec. Find the velocity of helicopter at the moment when its engine is switched off. Take velocity of sound 320 m/s.

50 \sqrt{2}

m/s

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60

m/s

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40 \sqrt{2}

m/s

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80

m/s

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Solution

The correct option is D $80$ m/s

Total time = $t_{1}$ + Time taken by sound to travel
$\Rightarrow 30 = t_{1} + \frac{\frac{1}{2} 3 t_{1}^{2}}{320}$
$(30 - t_{1}) 320 = \frac{3}{2} t_{1}^{2}$
$3 t_{1}^{2} + 640 t_{1} - 19200 = 0$
$t_{1} = 26.66$
$\Rightarrow v = 3 \times t = 3 \times 26.66 = 80 m / s$

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The correct option is D 80 m/s Total time = t1 + Time taken by sound to travel ⇒30=t1+123t21320 (30−t1)320=32t21 3t21+640t1−19200=0 t1=26.66 ⇒v=3×t=3×26.66=80 m/s