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Question about projectile motion

If a particle's vertical displacement from the origin is y and its horizontal displacement from the origin is x, is the direction of motion (angle to the horizontal) equal to

tan⁡−1dydx \tan ^{-1} \frac{dy}{dx}
?
gangsta316
If a particle's vertical displacement from the origin is y and its horizontal displacement from the origin is x, is the direction of motion (angle to the horizontal) equal to

tan⁡−1dydx \tan ^{-1} \frac{dy}{dx}
?



Yes.
Reply 2
Avatar for gangsta316
gangsta316
OP
13
Is there a specific reason to do with the definition of dy/dx? I got the trajectory equation and showed that it worked but is there a reason why? Can we apply this to curves and gradients?
Reply 3
Avatar for KSP
KSP
9
gangsta316
If a particle's vertical displacement from the origin is y and its horizontal displacement from the origin is x, is the direction of motion (angle to the horizontal) equal to

tan⁡−1dydx \tan ^{-1} \frac{dy}{dx}
?


If you mean the y coordinate over the x coordinate then yes, but not dy/dx.
gangsta316
Is there a specific reason to do with the definition of dy/dx? I got the trajectory equation and showed that it worked but is there a reason why? Can we apply this to curves and gradients?


Denoting y to represent vertical displacement, and x for horizontal displacement: dy/dt is vertical velocity, and dx/dt is horizontal velocity. So the angle to the horizontal is arctan[(dy/dt)/(dx/dt)] = arctan[dy/dx]. It's exactly the same with your standard Cartesian curves.

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