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carrotsss

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An object is projected downwards with initial velocity V0. The air resistance at speed v has magnitude mkv, where k is a positive constant. Take downwards as the positive





direction. 􏰃 (a) Show that t = 1 loge

k



􏰄

g−kV0 . g−kv





(b) Hence show that v = g (1 − e−kt) + V0 e−kt, and that the terminal velocity is g . k􏰀􏰁k





(c) Integrate again to show that x = gt+ kV0−g 1−e−kt . k k2





(d) Suppose that the terminal velocity of this object is 20 m/s, and that g = 10 m/s2. The object is thrown vertically downwards from a lookout at the top of a cliff at precisely the terminal velocity. At the same instant, a similar object is dropped from the same height. Show that the distance between the two falling objects after t seconds is

2
40(1 − e− 1 t) metres, and hence state the limiting distance between them.


uh copy paste didnt work but what does limiting distance mean? like is it max distance or what? that doesnt sound right but idk what it means?
limiting distance is what it approaches as time approaches infinity iirc
 

synthesisFR

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I thought its like
since one is dropped the terminal
then one is dropped from rest
as t goes to infiinity they both travel at terminal velcoity
then they are a constant distance between each other which is the distance that is travelled as the second particle accelerates to terminal
correct?
 

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