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?

Question

kkutie7 · Answer

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

kkutie7 · Answer

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

kkutie7 · Answer

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

anonymous · Answer

Thank you so much! :) @Kkutie7