Help? Kinematics: Projectile motion

A cannon is positioned on an inclined plane with angle alpha. Find angle beta. We want a projectile fired by cannon to hit inclined plane with angle 90°
see picture
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1 Answer
Mar 10, 2018

beta = tan^-1(1/2 cot alpha)

Explanation:

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