An aircraft is .flying. horizontally with a constant vefocity `=200m//s` at a height `=1km` above groun At the momement shown, a bomb is released from the aircraft and the cannon-gun below fires a shell with initial speed `=200m//s` , at some angle `theta` For what value of `theta` will the projectile shell destroy the bomb in mid-air? If the value of `theta` is `53^(3)` , find the minimum distance between the bomb and the shell as they fly past each other. Take `sin53^(@)=4//5`

Explanation

(a) Suppose the shell destroys the bomb at time `t`. Then for horizontal motion,
`t(200 + 200 cos theta) = sqrt(3) xx 1000`
`:. t(1 + cos theta) = 5 sqrt(3)`…(i)
For vertical motion,
`(1)/(2) "gt"^2 + (200 sin theta) t - (1)/(2) "gt"^2 = 1000 sin theta t = 5` ...(ii)
From (i) and (ii), `(sin theta)/(1 + cos theta) = (1)/(sqrt(3)`
On solving,we get `theta = 60^@`.
(b) `vec v_A = 200 i`
`vec v_B = -200 cos 53^@ hat i + 200 sin 53^@ hat j`
=` -200 xx (3)/(5) hat i + 200 xx (4)/(5) hat j = -120 hat i + 160 hat j`
`vec v_(A//B) = vec v_A - vec v_B = (200 + 120) hat i - 160 hat j`
`tan theta = |- (160)/(320)| = (1)/(2)`
`AB = 2 km`
`BP` = Miximum distance = `AB sin (30^@ - theta)`
`BP = 2 [sin 30^@ cos theta - cos 30^@ sin theta]`
=`2[(1)/(2) xx (2)/(sqrt(5))-(sqrt(5))/(2) xx (1)/(sqrt(5))] = (2 -sqrt(3))/(sqrt(5)) km`.
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