Terminal Speed in Projectile Motion under Air Resistance

AI Thread Summary
The discussion revolves around understanding terminal speed in projectile motion with air resistance, specifically at 43 m/s. The key equation involves the acceleration in the y-direction, factoring in gravitational acceleration and air resistance, represented by the term αvvy. Participants are clarifying whether to equate the vertical velocity vvy to the square of the terminal speed or to compute it separately. Additionally, there is a question about the behavior of the horizontal component of velocity when the object reaches terminal speed. The conversation highlights the complexities of analyzing motion under air resistance.
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Homework Statement


Terminal speed = 43 m s-1

Homework Equations


Acceleration in y direction when ball falling down: ay = - g + \alphavvy
where \alpha = K/m
and v = (vx2 + vy2)^0.5

The Attempt at a Solution


Putting ay = 0 as there is no acceleration when terminal speed occurs.
However, is vvy equal to v2, i.e. 432 here?
Or do I have to compute its velocity in the y direction?
 
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When at terminal speed (near as dammit), what can you say about the horizontal component?
 
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