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Standard XII

Physics

Applications of Horizontal and Vertical Components

A particle is...

Question

A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If $α$ and $β$ be the base angles and $θ$ the angle of projection prove that $tan θ = tan α + tan β$ .

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Solution

$tan α = \frac{y}{x}, tan β = \frac{y}{R - x} tan α + tan β = \frac{y R}{x (R - x)} y = \frac{x (R - x)}{R} (tan α + tan β) y = x tan θ (1 - \frac{x}{R}) tan θ = tan α + tan β$

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A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If α and β be the base angles and θ the angle of projection prove that tanθ=tanα+tanβ.

tanα=yx,tanβ=yR−xtanα+tanβ=yRx(R−x)y=x(R−x)R(tanα+tanβ)y=xtanθ(1−xR)tanθ=tanα+tanβ

A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If $α$ and $β$ be the base angles and $θ$ the angle of projection prove that $tan θ = tan α + tan β$ .

$tan α = \frac{y}{x}, tan β = \frac{y}{R - x} tan α + tan β = \frac{y R}{x (R - x)} y = \frac{x (R - x)}{R} (tan α + tan β) y = x tan θ (1 - \frac{x}{R}) tan θ = tan α + tan β$