A particle projected at a definite angle `prop` to the horizontal passes through points `(a,b) and (b,a)`, referred to horizontal and vertical axes through the points of projection. Show that :
(a) The horizontal range `R = (a^2 + ab + b^2)/(a + b)` .
(b) The angle of projection `prop` is given by
`tan^-1 [(a^2 + ab + b^2)/(ab)]`.

Options

A. `tan^(-1) [(P^2 +PQ+Q)/(PQ)]`
B. tan^(1) [(P^2+Q^2)/(PQ)]`
C. ` tan ^(-1)[(P^2 +Q^@)/(2 PQ)]`
D. ` sin ^(-1) [(P^2 +Q^2 + PQ)/(2 PQ)]`

Answer

A

Explanation

For angular projection of a projectile equation of trajectiory is
` y = x tan theta - ( gc^2)/( 2 u^2 cos^2 theta)`
` = x tan theta [ 1 - (gx)/(2 u ^2 2 theta) ] - x tan theta [ 1 - e/R]`
[ :. (u^2 sin 2 hteta 0/g = R = range of projectile ]`
as per question b,rgt ` Q = P tan theta [ 1- P/R]` ...(i)
and ` P = Q tan theta [ 1 - Q /R]` ...(i)
Diveding (i) by (ii), we get
` Q/P = P/Q ( R - P)/( R-Q) or Q^2 R- Q^3 r - P^3`
or ` R (P^2 - Q^1) = P^3 - Q^3`
or ` R = (P^3 -Q^3)/(P^2 -Q^1) = (P^2 + PQ+Q^2)/((P+Q)`
Putting this value in (i) we get
` ` Q= P tan hteta [ 1 - (P (P+Q))/(P^2+PQ+Q^2)]`
` = P tan theta [(P^2 +PQ+Q^2 -P^2 -PQ)/(P@+PQ+Q+Q^)]`
`= P tan hteta [ Q^2/( P^2 + PQ+Q^2)]`
or ` tan theta = (P^2 + PQ+q^2)/(PQ)`
or ` thea =tan^(-1) [(P^+PQ+Q^2)/(PQ)]` .