The horizontal range of a ground to ground projectile is `4sqrt(3)` times its maximum height. Its angle of projection is `pi/n` radian. Find n.

Answer

A-r, B-q, C-p, D-q

Explanation

[A-r] [B-r] [C-p] [D-q]
Given `y-x^(2)/(80)," comparing it with y"=x tan theta-(qx^(2))/(2u^(2) cos^(2) theta)" we have tan theta=1 rArr theta=45^(@)`
We have `(2u^(2) cos^(2) theta)/(g)=80 rArr (2u^(2) (1//2))/(10)=80. ("where "theta " angle of projection & u: velocity of projection")`
`rArr u^(2)=800 rArr u=20sqrt2m//s rArr u_(x)=u cos theta=20m//s and u_(y)=u sin theta =20m//s`
`rArr" after 4 sec ", v_(y)=u_(y)+a_(y)t=20-40=-20m//s`
`rArr` angle made by velocity vector with horizontal `=tan^(-1)""(v_(y))/(u_(x))=45^(@)` below horizontal Maximum height `=(u_(y)^(2))/(2g)=20m" Horizontal range "=(2u_(x)u_(y))/(g)=(2 xx 20 xx 20)/(10)=80`