Timeline for Vertical movement of an object with gravitational acceleration not constant
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2022 at 20:30 | vote | accept | CommunityBot | ||
| Dec 15, 2022 at 19:47 | comment | added | user353361 | Oh, I understand know! Thank you very much for your answer, really helpful! | |
| Dec 15, 2022 at 19:33 | history | edited | Eli | CC BY-SA 4.0 |
added 612 characters in body
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| Dec 15, 2022 at 19:27 | history | edited | Eli | CC BY-SA 4.0 |
added 612 characters in body
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| Dec 15, 2022 at 16:33 | comment | added | Eli | I see, which data you used $~v_0~,R~,g_0~$ ? | |
| Dec 15, 2022 at 15:46 | comment | added | user353361 | Which is $15m$ more from what will you obtain if you consider that all along the way, while it rises the initial gravitational acceleration remains constant. I also know that the difference in the results is insignificant for the huge velocity that I took into account and that for small velocities which are usually worked with, the difference is more insignificant. Does it have a logic because I'm in early highschool and I really want to understand physics? Nevertheless, thanks for the answer. | |
| Dec 15, 2022 at 15:45 | comment | added | user353361 | Yeah, I think that I figured it out somewhat. I wrote $$g_f = \frac{g_0 \cdot \left(3 \cdot h_f^2 - 2 \cdot R \cdot h_f + R^2 \right)}{R^2} \text{. Used binomial expansion!}$$ $$g_{average} = \frac{g_0 \cdot \left(3 \cdot h_f^2 - 2 \cdot R \cdot h_f +2 \cdot R^2\right)}{2 \cdot R^2}$$ $$\frac{m \cdot v_0^2}{2} = m \cdot g_{average} \cdot h_f \implies \frac{v_0^2 \cdot R^2}{g_0} = 3 \cdot h_f^3 - 2 \cdot R \cdot h_f^2 +2 \cdot R^2 \cdot h_f$$. Above we have a cubic equation(I know that it's strange), where $h_f$ has 2 imaginary solutions and 1 real, the real one is $h_f \approx 10015.48178 m$ | |
| Dec 14, 2022 at 16:46 | history | answered | Eli | CC BY-SA 4.0 |