Question

A projectile is projected up an inclined plane of inclination alpha at an elevation theta to the horizontal .Show that tantheta=cotalpha+2tanalpha

Answer 1

When the particle’s velocity is horizontal it is at maximum height HH and has travelled a horizontal distance LL equal to half of the maximum range (R/2R/2).

H=u2sin2α2gH=u2sin2⁡α2g

L=R2=u2sin2θ2gL=R2=u2sin⁡2θ2g

tanβ=HL=u2sin2α/2gu2sin2α/2g=tanα2tan⁡β=HL=u2sin2⁡α/2gu2sin⁡2α/2g=tanα2

tanα=2tanβtanα=2tan⁡β

hope it helps...