# Thread: Motion in a resisting medium

1. ## Motion in a resisting medium

A body of mass m is projected vertically upwards with speed u. Air resistance is equal kv^2. Find the speed when next at the point of projection.

2. Originally Posted by dtugiono
A body of mass m is projected vertically upwards with speed u. Air resistance is equal kv^2. Find the speed when next at the point of projection.
Can you set up the differential equation for the upwards motion?

Can you set up the differential equation for the downwards motion?

Please post what you've done so far.

3. dv/dt=-g-kv^2
I don't know what to do next?

4. Originally Posted by dtugiono
dv/dt=-g-kv^2
I don't know what to do next?
$\displaystyle \frac{dv}{dt} = - g - \frac{k}{m} v^2$

$\displaystyle \Rightarrow \frac{dt}{dv} = - \frac{1}{g + \frac{k}{m} v^2} = - \frac{m}{mg + k v^2}$ subject to the boundary condition that v = u at t = 0.

Solve this differential for t as a function of v and then make v the subject. Get x from v and use it to find x when v = 0 (the maximum height).

Now set up and solve the differential equation for the downwards motion. Find v when x = distance found above.

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