At a point P on a ground-to-ground projectile, the line joining P to the point of projection makes angle α with the horizontal and the line joining P to the point of landing makes acute angle β with the horizontal. If ψ is the angle made by the tangent to the trajectory at P with the horizontal, then:
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Question

At a point P on a ground-to-ground projectile, the line joining P to the point of projection makes angle α with the horizontal and the line joining P to the point of landing makes acute angle β with the horizontal. If ψ is the angle made by the tangent to the trajectory at P with the horizontal, then:

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G1 Physics · Accepted Answer

For the trajectory y = x tanθ(1−x/R), the slope of the tangent is dy/dx = tanθ(1−2x/R). From the geometry of point P, tanα = tanθ(1−x/R), tanβ = tanθ(x/R). Therefore tanα−tanβ = tanθ(1−2x/R) = tanψ. Hence tanψ=tanα−tanβ.