How do I calculate this projectile motion?

Question

A projectile is launched with initial speed v (and angle α to the horizontal) at the base of a plane inclined at an angle θ to the horizontal. Show that when it hits the plane,

x = 2v2/g cos2α (tan α – tan θ)

= v2/g cos θ [sin(2 α – θ) - sin θ)

Answer 1

Let us assume that the projectile hits the plane at time t after being launched.

Horizontal component of velocity = vcosα

There is no force in horizontal direction. Therefore, horizontal component of velocity is constant.

Therefore, horizontal displacement x = (vosα)*t = vtcosα----(1)

Vertical component of initial velocity = vsinα

Vertical acceleration =-g

Vertical displacement y = (vsinα)t - 1/2*g*t^2-----(2)

Dividing (2) by (1)

y/x = {(vsinα)t - 1/2*g*t^2}/(vtcosα)

But y/x = tan θ

Therefore,

{(vsinα)t - 1/2*g*t^2}/(vtcosα) = tan θ

Or, (vsinα - 1/2 gt)/(vcosα) = tan θ

Or, vsinα - (1/2) * gt = vcosα * tan θ

Or, gt/2 = vsinα - vcosα * tan θ

Or, t = 2*(vsinα - vcosα * tan θ)/g----------------(3)

Using (3) in (1)

x = v*2*{(vsinα - vcosα * tan θ)/g} *cosα

Or, x = 2v^2/g * (sinα - cosα * tanθ)*cosα

Can you do the rest?