Question

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

Answer 1

`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`