# Mechanics Question? (1 Viewer)

#### Lalji

##### New Member
Point P, Q are h metres apart in a vertical line. A particle is projected with speed C from P towards Q at the same instant, another is projected with speed C from Q towards P.

If the resistance of the medium is proportional to the speed and the terminal speed is C, show that the particles meet after time

$\frac{C}{g}\log_e\left(\frac{2C^2}{2C^2-gh}\right)$

provided $2C^2>gh$.

As per upwards and downwards cases

I could get (if I am right!)
for upwards:

$t_1=\frac{1}{k} \log_e \left(\frac{kC+g}{kv+g}\right)$
$x_1=\frac{1}{k}\left[C-v+\frac{g}{k}\log_e \left(\frac{kv+g}{kC+g}\right)\right]$
for downwards:
$t_2=\frac{1}{k} \log_e \left(\frac{g-kC}{g-kv}\right)$
$x_2=\frac{1}{k}\left[C-v+\frac{g}{k}\log_e \left(\frac{g-kC}{g-kv}\right)\right]$

I understand
$t_1=t_2$
$x_1+x_2=h$

How do I use terminal velocity C to get the result?

Using your equations for time, make $v$ the subject. Integrate again to get $x$ as a function of $t$, and then solve $x_1 + x_2 = h$.
Note that $C = g/k \implies k = g/C$.