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The angular elevation of an enemy's posi...

The angular elevation of an enemy's position on a hill 'h' ft height is `alpha`.Wha should be the minimum velocity of the projectile in order to hit the enemy?

A

`u=sqrt(gh(cos alpha+1))`

B

`u=sqrt(gh(sin alpha+1))`

C

`u=sqrt(gh(cossec alpha+1))`

D

`u=sqrt(gh(sec alpha+1))`

Text Solution

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The correct Answer is:
C
`O` is the point of projection of the shell and `A` is the position of enemy at a height `h` above the level of `O`.
If `u` is the minimum initial velocity of the projectile to shell the enermy, then `OA` must be the maximum range up the inclined plane of angle `alpha`.
So `OA=u^(2)/(g(1+sin alpha))`....(i)
From `DeltaOAB, OA= h cosec alpha`....(ii)
From eqn.(i) and (ii) , `u=sqrt(gh(cosec alpha+1))`
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