Calculating Trajectory Equation of Motion for a Rock Touching a Wall

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PeroK said:
It was even simpler to have the origin on the parabola itself at the top of the first wall, so that ##c = 0##
Yes that is a better choice.
 
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compaq65 said:
Why there is a ##c## constant, if in
$$y = x\tan \theta - \frac{gx^2}{2v^2}\sec^2 \theta$$
I can only notice ##a## and ##b## coefficients. Because of that I get a little bit different answers. Which method is correct?
You should realize by now that this problem has nothing to do with the kinematics of the projectile. Velocity and acceleration are irrelevant. Your answer does not depend on them. The same question can be asked about a parabola painted on a wall.
 
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Everything is clear now. Thanks, you all helped me a lot!
 
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