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Grade: 11
        
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
2 years ago

Answers : (1)

Ashutosh Mohan Sharma
askIITians Faculty
180 Points
							509-636_CF85C169-BCCA-4776-9160-52BAA602F05C.jpegGiven 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
2 years ago
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