A projectile is fired from the base of a cone-shaped hill. The projectile grazes the vertex and strikes the hill again at the base. If alpha be the half angle of the cone, then the angle of projection is

Given that a particle is thrown withangle of projectionθover a triangleΔACBfrom one of its endAof the horizontal baseABaligned along X-axis and it finally falls at the other endBof the base,grazing the vertexC(x,y)

Letube the velocity of projection,Tbe the time of flight,R=ABbe the horizontal range andtbe the time taken by the particle to reach at C(x,y)

The horizontal component of velocity of projection→ucosθ

The vertical component of velocity of projection→usinθ

Considering motion under gravity without any air resistance we can write

y=usinθt−12gt2.....[1]

x=ucosθt...................[2]

combining [1] and [2] we get

y=usinθ×xucosθ−12×g×x2u2cos2θ

⇒y=usinθ×xucosθ−12×g×x2u2×sec2θ

⇒yx=tanθ−(gsec2θ2u2)x........[3]

Now during time of flightTthe vertical displacement is zero
So

0=usinθT−12gT2

⇒T=2usinθg

Hence horizontal displacement during time of flight i.e. range is given by

R=ucosθ×T=ucosθ×2usinθg=u2sin2θg

⇒R=2u2tanθg(1+tan2θ)

⇒R=2u2tanθgsec2θ

⇒gsec2θ2u2=tanθR......[4]

Combining [3] and [4] we get

yx=tanθ−12×gxu2×sec2θ

⇒yx=tanθ−xtanθR

⇒tanα=tanθ−xtanθR[sinceyx=tanαfrom figure]

Sotanθ=tanα×(RR−x)

⇒tanθ=tanα×(R−x+xR−x)

⇒tanθ=tanα×(1+xR−x)

⇒tanθ=tanα+xtanαR−x

⇒tanθ=tanα+yR−x[puttingxtanα=y]

Finally we have from figureyR−x=tanβ

Hence we get our required relation

tanθ=tanα+tanβ