# Perpendicular Projectile

• Carbon123
In summary, the conversation discusses a projectile being launched at a height and angle from the bottom of an inclined plane. The question is how to determine the velocity of the projectile so that it strikes the plane perpendicularly. One solution involves setting the x-component of velocity to zero, while another solution involves using normal coordinates and setting the ratio of the x and y components to the tangent of the angle.

## Homework Statement

A projectile is launched at height h with velocity v with 0 degrees angle with respect to the ground from the bottom of an inclined plane with an angle of theta.Determine the particle's velocity so that it strikes the inclined plane perpendiculary.

y=Vo*t+1/2at^2

## The Attempt at a Solution

Could anyone help me to check m work and my concepts ?
First ,I did it by changing the x coordinates to the distance along the inclined plane,then the y coordinates are perpendicular to the inclined plane.
So,my question is ,when it strikes the plane perpendicularly,will it have 0 Vx ?[/B]

If the x-axis is along the plane, then, yes, the x-component of velocity must be zero for the projectile to strike the plane perpendicularly.
In the proposed scheme, you realize that gravity will point at an angle to the y axis?

Yes, the x-component of the projectile's velocity will be zero when it crashes perpendicular into the plane.
What should the angle of its velocity then be in the "normal\usual" x-y coordintate system?

@SimonBridge Yes i solved the problem by setting Vx in the changed coordinates to 0,and also have g equal g cos theta in y direction,and g sin theta in x direction.@andrevdh So if it is in normal coordinates then it will have Vx /Vy =tan theta (probably ?).I haven't tried to use normal coordinates yet.