A particle at a height `h` from the ground is projected with an angle `30^@` from the horizontal, it strikes the ground making angle `45^@` with horizontal. It is again projected from the same point with the same speed but with an angle of `60^@` with horizontal. Find the angle it makes with the horizontal when it strikes the ground

Options

A. `tan^(-1) (4)`
B. `tan^(-1)(5)`
C. `tan^(-1)(sqrt(5))`
D. `tan^(-1)(sqrt(3))`

Answer

3

Explanation

vertical velocity when it reaches ground
`=v^(2)=u^(2)+2gh " " rArr u^(2) sin phi+2gh `
corresponding horizontal velocity `rArr u cos theta`
`(u^(2)sin^(2)30+2gh)/(u^(2) cos^(2) 30)=tan 45=1`
`rArr u=2sqrt(gh)`
`tan theta=(u^(2) sin^(2)60+2gh)/(u^(2)cos^(2)60)=sqrt(5)`