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`        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

Ashutosh Mohan Sharma
180 Points
```							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→ucosθThe vertical component of velocity of projection→usinθConsidering motion under gravity without any air resistance we can writey=usinθt−12gt2.....[1]x=ucosθt...................[2]combining [1] and [2] we gety=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 zeroSo0=usinθT−12gT2⇒T=2usinθgHence horizontal displacement during time of flight i.e. range is given byR=ucosθ×T=ucosθ×2usinθg=u2sin2θg⇒R=2u2tanθg(1+tan2θ)⇒R=2u2tanθgsec2θ⇒gsec2θ2u2=tanθR......[4]Combining [3] and [4] we getyx=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=tanβHence we get our required relationtanθ=tanα+tanβ
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2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions