# A particle is dropped from the top of a tower h meter high and simultaneously a particle is projected upwards from bottom. They meet when the upper one has descended 1/n of the distance. find the initial velocity of the lower particle?

Jun 23, 2015

The initial velocity of the lower particle was ${v}_{0} = \sqrt{\frac{n g h}{2}}$

#### Explanation:

So, you know that you're dealing with two particles, one at the top of a tower that's h meters high, and the other one on the bottom.

Moreover, you know that if you drop the top particle with no initial velocity and launch the bottom particle with an initial velocity equalto ${v}_{0}$, they meet when the upper particle travelled $\frac{1}{n}$ of the heigt of the tower.

For the upper particle, the distance it travelled can be written as

$\frac{1}{n} \cdot h = \underbrace{{v}_{0 \text{upper"))_(color(blue)("=0}}} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

$\frac{1}{n} \cdot h = \frac{1}{2} \cdot g \cdot {t}^{2}$ $\text{ } \textcolor{b l u e}{\left(1\right)}$, where

$t$ - the time needed for the particle to reach that height.

The bottom particle will travel a distance of

$h - \frac{1}{n} \cdot h = \frac{\left(n - 1\right)}{n} \cdot h$

For the bottom particle you can write

$\frac{\left(n - 1\right)}{n} \cdot h = {v}_{0} \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}$ $\text{ } \textcolor{b l u e}{\left(2\right)}$

Since the two particles meet up at this point, the time of flight for both particles must be equal.

Use equation $\textcolor{b l u e}{\left(1\right)}$ to find a relationship for ${t}^{2}$

${t}^{2} = \frac{2 h}{n g}$

Plug this into equation $\textcolor{b l u e}{\left(2\right)}$ to get

$\frac{\left(n - 1\right)}{n} \cdot h = {v}_{0} \cdot \sqrt{\frac{2 h}{n g}} - \frac{1}{\cancel{2}} \cdot \cancel{g} \cdot \frac{\cancel{2} h}{n \cancel{g}}$

$\frac{\left(n - 1\right)}{n} \cdot h = {v}_{0} \cdot \sqrt{\frac{2 h}{n g}} - \frac{h}{n}$

Rearrange to isolate ${v}_{0}$ on one side of the equation

${v}_{0} \cdot \sqrt{\frac{2 h}{n g}} = \frac{\left(n - 1\right)}{n} \cdot h + \frac{h}{n}$

${v}_{0} \cdot \sqrt{\frac{2 h}{n g}} = \frac{\cancel{n}}{\cancel{n}} \cdot h - \cancel{\frac{h}{n}} + \cancel{\frac{h}{n}}$

${v}_{0} = \frac{h}{\sqrt{\frac{2 h}{n g}}} = \frac{\cancel{h} \cdot \sqrt{\frac{2 h}{n g}}}{\frac{2 \cancel{h}}{n g}} = \frac{n g}{2} \cdot \sqrt{\frac{2 h}{n g}}$

This is equivalent to

v_0 = sqrt( (n^(cancel(2)) * g^(cancel(2)))/4 * (2h)/((cancel(n) * cancel(g))

${v}_{0} = \textcolor{g r e e n}{\sqrt{\frac{n g h}{2}}}$