A particle is projected from a point `P(2,0,0) m` with a velocity `10 m//s` making an angle `45^(@)` with the horizontal. The plane of projectile motion passes through a horizontal line `PQ` which makes an angle of `37^(@)` with positive `x`-axis, `xy` plane is horizontal. the coordinates of the point where the particle will strike the line `PQ` is : (Take `g=10 m//s^(2)`)

Options

A. `(10, 6, 0) m`
B. `(8, 6, 0)m`
C. `(10, 8, 0) m`
D. `(6, 10, 0)m`

Answer

A

Explanation

`PQ=R=(u^(2) sin 90^(@))/(g)=(100xx1)/(10)=10 rArr PQ=10`

`(x,y)=(10, 6, 0)`