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March 21st, 2017, 11:55 AM   #1
Joined: Jun 2015
From: Warwick

Posts: 37
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A particle projected in an upwards direction with resistance $\frac{gv}{V}$.

A particle is projected upwards in a medium whose resistance is $\frac{gv}{V}$, where $v$ is the velocity. If $V$ is large compared to $U$, which is the velocity of projection, show that the fraction of the value of the vertical height $h$ reached by the particle is described as $$\frac{2U}{3V}$$ and the fraction of the value of the ascent time is $$\frac{U}{2V}$$ and the fraction of the valure of the decent time is $$\frac{U}{6V}$$ when there is no resistance.

$V$ can be taken as the terminal velocity of the particle, although the question does not state this. The particle could reach the terminal velocity on its way upwards, before its return journey.

Prove that the particle returns to the point of projection with the velocity $$U(1- \frac{2U}{3V})$$

Can anyone derive the above results from the projectile equation for projecting a particle vertically upwards or would you set up and solve a differential equation?

Last edited by Statistics132; March 21st, 2017 at 12:04 PM.
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