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A package is dropped form an airplane travelling horizontally at constant speed. Friction is negligible. One second later a second package is dropped.

a) THe distance between the two packages will remain constant as they fall b) The distance between the two packages will steadily increase as they fall c) The second package will hit the ground more than one second after the first hits d) The horizontal distance between the two packages will decrease as they fall.

The answer is b, can someone explain why?

I thought that answer is going to be a, because it makes logical sense for me.

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closed as off-topic by Sebastian Riese, John Rennie, DanielSank, user36790, Kyle Kanos Oct 20 '15 at 10:18

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For the first parcel, the distance fallen as a function of time is $d_1=(g/2)t^2$. Note the quadratic dependence on $t$. The 2nd parcel is dropped at $t=1s$. This means the distance fallen for the second parcel is $d_2=(g/2)(t+1)^2=(g/2)(t^2+2t+1)$. So the change in distance for the two parcels (the horizontal distance remains equal due to gravity being a vertical force and the plane traveling at a constant velocity) is $d=d_2-d_1=(g/2)(t^2+2t+1)-(g/2)t^2=(g/2)(2t+1)$, which increases with time. Hence the distance increases. I hoped this helped answer your question!

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