The discussion centers on deriving the trajectory equation of motion for a rock touching a wall, utilizing equations such as $$x(t)=v\cos\alpha t$$ and $$y(t)=v\sin\alpha t-\frac{1}{2}gt^{2}$$. Participants explore various relationships between the variables, including $$h=xtan\alpha -\frac{x^{2}g(1+tan^{2}\alpha )}{2v^{2}}$$ and $$nr=\frac{v^{2}\sin2\alpha }{g}$$. The challenge lies in solving a system of equations with four equations and five unknowns, leading to discussions on simplifying the problem by analyzing the middle section of the trajectory and eliminating variables.
PREREQUISITESStudents and professionals in physics, engineers working on projectile motion problems, and anyone interested in the mathematical modeling of trajectories.
Yes that is a better choice.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##
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.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?