Timeline for Conceptual question regarding velocity in projectile motion vs. up a ramp
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2019 at 5:00 | history | edited | TechDroid | CC BY-SA 4.0 |
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| Mar 9, 2019 at 21:29 | comment | added | BioPhysicist | I guess you are just using words differently than I am used to. You say things like "vertical kinetic energy", and you seem to say that the OP is wrong in saying that the final kinetic energy is $\frac12m(v\cos\theta)^2$ when this is in fact the kinetic energy of the projectile at maximum height. There is no energy conservation broken here. | |
| Mar 9, 2019 at 21:08 | comment | added | TechDroid | I never said there was, but if they're not distinguished or categorized, things become complicated. You can see the OP runs into problem with the first case because his final kinetic energy is derived from the horizontal velocity of the object while the vertical velocity is zero. This definitely is something that needs to be contrasted so OP will understand which velocity is being transformed to potential at the apogee. | |
| Mar 9, 2019 at 20:30 | comment | added | BioPhysicist | Your very first sentence says they broke conservation of energy. Energy isn't a vector. There isn't horizonal and vertical energy | |
| Mar 9, 2019 at 18:19 | history | edited | TechDroid | CC BY-SA 4.0 |
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| Mar 9, 2019 at 18:13 | comment | added | TechDroid | @ Stevens. My answer didn't go against that, but since the OP is in search of potential energy at the max height, the horizontal energy of the body shouldn't come in anymore. | |
| Mar 9, 2019 at 17:50 | comment | added | BioPhysicist | The work done by the OP is correct | |
| Mar 9, 2019 at 7:07 | history | answered | TechDroid | CC BY-SA 4.0 |