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  • A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight when the target is seen, should the pilot drop the bomb in order to attack the target?

A fighter plane is flying horizontally at an altitude of 1.5 km with a speed of 720 km/h. At what angle of sight when the target is seen, should the pilot drop the bomb to attack the target?

 

Answers (1)

Given :u=720\; km/h=200\; m/s

Let us assume that the bomb is dropped by the pilot t seconds vertically above Q before the target T vertical component velocity of the bomb will be zero and the horizontal component value will be equal to that of the plane. So, the bomb covers the distance TQ as a free fall.

Now, u=0,H=1.5,g=10

So, H=ut+\frac{1}{2}gt^{2}=0+\frac{1}{2}10\times t\times t=1500

So, we get, t=10\sqrt{3}\; s

ut is the distance covered by the plane or the bomb equal to PQ

hence, PQ=ut=200\sqrt{3}\; m

now, tan\theta=\frac{TQ}{PQ}=\sqrt{\frac{3}{4}}

Hence, angle of sight \theta=23^o

Posted by

infoexpert23

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