Get Answers to all your Questions

header-bg qa

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?

 

Answers (1)

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

we 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 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 plane or the bomb equal to PQ

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

now, \tan \theta =\frac{TQ}{PQ}=\sqrt{\frac{3}{4}}=23^{o}42'

Posted by

infoexpert23

View full answer