Help? Kinematics: Projectile motion

It is easiest to solve this problem if we set up an coordinate system with the origin at the point of projection, the #X# axis directed up the plane, and the #Y# axis perpendicular to the plane. Thus, if the speed of projection is #u#, the components of the initial velocity are #u_x = u cos beta# and #u_y=u sin beta#.

In this coordinate system, both the #X# and #Y# components of the motion are accelerated, with the accelerations being

#a_x =-g sin alpha, qquad a_y = -g cos alpha#

Quick check: this reduces to the familiar #a_x=0,quad a_y =-g# for the case #alpha=0#

Thus we have

#v_x = u_x+a_x t = u cos beta -g sin alpha quad t#
#v_y = u_y+a_y t = u sin beta -g cos alpha quad t#

and

#x = u cos beta quad t -1/2 g sin alpha quad t^2#
#y = u sin beta quad t -1/2 g cos alpha quad t^2#

According to the problem, when the projectile hits the plane on its way down (#y=0#), we must have #v_x=0#. But putting #y = 0# gives the time of flight as

#t = {2 u sin beta}/{g cos alpha}#

while #v_x=0# gives

#t = {u cos beta}/{g sin alpha}#

Equating these two values for #t# leads to

#{2 u sin beta}/{g cos alpha} = {u cos beta}/{g sin alpha} implies tan beta = 1/2 cot alpha#