Analyzing the kinematic x(t) equation

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SUMMARY

The forum discussion centers on solving the kinematic equation deltaY = (tan A)deltaX - g*deltaX^2/2V(initialx)^2 for deltaX symbolically. Participants analyze the relationship between the terms, particularly questioning the absence of time (t) in the equation and the role of gravitational acceleration (g) in relation to initial velocity (V(initial)). The discussion highlights the dimensional inconsistency of the equation and clarifies that V(initial) cannot simply be represented as tan A.

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Homework Statement



Solve the equation below for deltaX symbolically and simplify the answer as much as possible.

deltaY = (tan A)deltaX - g*deltaX^2/2V(initialx)^2

Homework Equations


quadratic equation.


The Attempt at a Solution



I can handle the algebra and simplification. My question comes as I analyze the equation as a kinematics problem. I assume the fnal term was gotten by the equation:

V(final)^2 = V(initial)^2 + 2ax. Therefore V(final) = 0 so there is a deceleration. But g is divided by this accel so I guess g is dependent on this acceleration??

I also found, that given the velocity v time graph, V(initial) = tan A.

Why wasn't this value used in the g*deltaX^2/2 V(initialx)^2 term?

Finally, because I don't see any t in the equation, I assume it can be used in terms of deltaX as well?? Or was the professor just testing our understanding of the quadratic equ?

Does this quation describe any familiar motion?
 
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deltaY = (tan A)deltaX - g*deltaX^2/2V(initialx)^2
Dimensionally this equation is not correct. One t is missing.
I also found, that given the velocity v time graph, V(initial) = tan A.
V initial cannot be tanA
 

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