Projectile Velocity for Perpendicular Impact on an Inclined Plane

AI Thread Summary
To determine the velocity of a projectile striking an inclined plane perpendicularly, the x-component of the velocity must be zero at impact. The discussion confirms that when the projectile strikes the plane perpendicularly, gravity's components must be adjusted to account for the incline, with g acting at angles to the usual axes. The relationship between the velocity components in normal coordinates is suggested to be Vx/Vy = tan(theta). The participant successfully solved the problem by setting Vx to zero in the transformed coordinate system. Overall, the key focus is on understanding the velocity components relative to the inclined plane during perpendicular impact.
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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.

Homework Equations


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]
 
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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? :wink:
 
@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.
 
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