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I am trying to work through a problem, that is basically telling me that two objects were dropped from a building, one was thrown 2 seconds after the previous object, and I have to find out how long it takes for these two objects to be a certain length apart (15 meters for example).

My first guess at this was to use the $y = vt + 0.5at^2$ formula, which would give me the distance the first object travelled before the second object was thrown. I am not sure where to go from here.

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closed as off-topic by John Rennie, Neuneck, JamalS, Kyle Kanos, ACuriousMind Oct 8 '14 at 11:48

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  • $\begingroup$ Why dont you apply that equation to the second object too? $\endgroup$ – soumyadeep Oct 8 '14 at 6:56
  • $\begingroup$ The problem is I am not sure where to go after calculating the distance of the first object after 2 seconds. If I applied the equation to the second object wouldn't the answer be the same as the first object when I calculated it? $\endgroup$ – user3247128 Oct 8 '14 at 6:59
  • $\begingroup$ But the $t$ in the second one will be $t-2$ $\endgroup$ – soumyadeep Oct 8 '14 at 7:01
  • $\begingroup$ Thank you for your help I really do appreciate it. I am just very confused still although I kind of understand what you mean.... the first object I did; Y = 0 + (9.80 m/s^2)(2)^2....how do you mean change the second to t - 2 -? $\endgroup$ – user3247128 Oct 8 '14 at 7:06
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Your approach is correct, now simply add indices to everything, i.e.

$$y_i = v_{0,i}t_i + \frac12 a_it_i^2\quad\text{where } i\in\{1,2\}$$

and note that $t_2 = t_1 - 2\,\text{s}$. Then solve $15\,\text{m}\stackrel!=y_1 - y_2$.

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    $\begingroup$ I appreciate this thank you! I have never done physics before and this is my first week, so I am honestly clueless with it. I am trying to understand what you mean so thank you $\endgroup$ – user3247128 Oct 8 '14 at 7:14
  • $\begingroup$ You're welcome - it might be nice to add some information on your background to the question or your user profile, e.g. whether you're on high school, know about indices and the like etc. $\endgroup$ – Tobias Kienzler Oct 8 '14 at 7:26
  • $\begingroup$ I need to do that. I am in University but never took a physics class in high school! What are y1 and y2? Thank you! $\endgroup$ – user3247128 Oct 8 '14 at 7:46
  • $\begingroup$ $y_1$ and $y_2$ describe the position of objects 1 and 2 respectively, and the same applies to the initial velocities $v_{0,1}$ and $v_{0,2}$ (which vanish in your case), accelerations $a_1=a_2=g$ and times $t:=t_1=t_2-2\,\text{s}$. In general, you label (or enumerate) multiple objects and add the same label to their variables (coordinates etc) $\endgroup$ – Tobias Kienzler Oct 8 '14 at 7:49
  • $\begingroup$ you're welcome. Don't forget you can mark an answer as accepted if it solved your problem. And welcome to physics.SE! If you like it here, please take the tour for a quick introduction, and for formulae you'll definitely enjoy using LaTeX ;) $\endgroup$ – Tobias Kienzler Oct 8 '14 at 7:58
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If you take the time of throwing of the first object as $0$ then the second object will start falling at $0+2$ second.Now for the second one take the time of its start of fall as 0 and so the ending time will be 2 sec less i.e$0+2 \longrightarrow 0$ and$t \longrightarrow t-2$

So for the second one $$ \frac {dx} {dt} =u+at$$ $$ dx=udt +atdt$$ $$\int^y_0 dx=\int^{t-2}_0 udt+atdt$$

After you integrate with this limits your second equation will come out with $t-2$ .

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  • $\begingroup$ ahh I see what you mean now thank you for explaining it :) $\endgroup$ – user3247128 Oct 8 '14 at 7:45
  • $\begingroup$ And the answer you accepted is wrong it will be $t-2$ not $t+2$ $\endgroup$ – soumyadeep Oct 8 '14 at 8:20
  • $\begingroup$ @soumyadeep That depends on whether you label the first particle to be dropped as "1" or the upper one ;) but you're right, I fixed that $\endgroup$ – Tobias Kienzler Oct 8 '14 at 8:26

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