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Physics 8 Online
OpenStudy (anonymous):

Two bodies situated at different height from ground starting falling freely under gravity. The times of the fall of the two bodies are 1 s and 2 s respectively , If g=9.8 m/s2 , The the initial vertical separation between the two bodies is?

OpenStudy (kkutie7):

well lets start by drawing a picture:|dw:1453007933141:dw|

OpenStudy (kkutie7):

It's been a long while for me so check me stuff before you trust me

OpenStudy (kkutie7):

use the equation: \[y=y_{0}+v_{0}t+\frac{1}{2}gt^{2}\] y will be the ending position 0 and initial velocity will also be zero because we assume both bodies started at rest. B1: \[0=y_{0}+\frac{1}{2}(\frac{9.8m}{s^{2}})(1s)^{2}\] B2: \[0=y_{0}+\frac{1}{2}(\frac{9.8m}{s^{2}})(2s)^{2}\] you can now solve for the initial positions and compare them with a little algebra

OpenStudy (anonymous):

Thank you so much! :) @Kkutie7

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