Question

If the equation of trajectory of a particle in vertical plane is y = ax-bx ^1/2 where a and b are positive constants then angle of elevation of highest point from the point of projection is
(1)tan^-1 (a )
(2)tan^-1 (a/2)
(3)tan^-1 (b)
(4)tan^-1 (b/2)​

Answer 1

correct equation of trajectory of a particle in vertical plane is

y = ax - bx²

at start and end point of trajectory

ax - bx² = 0

x (a - bx) = 0

x = 0  , x = a/b

y will be maximum in middle

x = a/2b

y(a/2b) = a(a/2b) - b(a/2b)² = a²/4b

y max = a²/4b  at x = a/2b

Angle of elevation of highest point from point of projection = θ

tanθ = (a²/4b)/(a/2b)

=> tanθ = a/2

=> θ = tan⁻¹(a/2)

option 2 is correct