- Thread starter
- #1

#### DrunkenOldFool

##### New member

- Feb 6, 2012

- 20

- Thread starter DrunkenOldFool
- Start date

- Thread starter
- #1

- Feb 6, 2012

- 20

- Jan 26, 2012

- 890

The two horizontal components of velocity are constant, and the vertical components are always equal. Now draw a velocity diagram for when the two velocities are perpendicular, and solve the diagram for the vertical component of velocity (it should come to \(2 \sqrt{3}\) m/s if my scratch algebra and arithmetic are correct).

Now you can find the time \(t\) it took the vertical component to reach this value, and the seperation is \(7t\) m.

CB

Last edited:

- Jan 27, 2012

- 95

Let the velocities of the particles (say $\vec{v_{1}}'$ and $\vec{v_2 }'$) become perpendicular after time $t$. By equation of motion,

$$ \vec{v_{1}'}=\vec{v_{1}}+\vec{g}t \\ \vec{v_{2}'}=\vec{v_{2}}+\vec{g}t$$

As $\vec{v_1 ' }$ and $\vec{v_2 '}$ are perpendicular, we can write

$$ \begin{align*} \vec{v_1 ' } \cdot \vec{v_2 ' } &=0 \\ (\vec{v_{1}}+\vec{g}t) \cdot (\vec{v_{2}}+\vec{g}t) &= 0 \\ -v_1 v_2 +g^2 t^2 &=0 \\ t &= \frac{\sqrt{v_1 v_2}}{g}\end{align*}$$

Let $x_1$ and $x_2$ be the horizontal distances covered by particles 1 and 2 in time $t$ respectively. Note that the acceleration in horizontal direction is zero.

$$x_1 = v_1 t = v_1 \frac{\sqrt{v_1 v_2}}{g} \\ x_2 = v_2 t = v_2 \frac{\sqrt{v_1 v_2}}{g}$$

The total separation between the particles is

$$x_1+x_2= (v_1+v_2)\frac{\sqrt{v_1 v_2}}{g}$$

$$ \vec{v_{1}'}=\vec{v_{1}}+\vec{g}t \\ \vec{v_{2}'}=\vec{v_{2}}+\vec{g}t$$

As $\vec{v_1 ' }$ and $\vec{v_2 '}$ are perpendicular, we can write

$$ \begin{align*} \vec{v_1 ' } \cdot \vec{v_2 ' } &=0 \\ (\vec{v_{1}}+\vec{g}t) \cdot (\vec{v_{2}}+\vec{g}t) &= 0 \\ -v_1 v_2 +g^2 t^2 &=0 \\ t &= \frac{\sqrt{v_1 v_2}}{g}\end{align*}$$

Let $x_1$ and $x_2$ be the horizontal distances covered by particles 1 and 2 in time $t$ respectively. Note that the acceleration in horizontal direction is zero.

$$x_1 = v_1 t = v_1 \frac{\sqrt{v_1 v_2}}{g} \\ x_2 = v_2 t = v_2 \frac{\sqrt{v_1 v_2}}{g}$$

The total separation between the particles is

$$x_1+x_2= (v_1+v_2)\frac{\sqrt{v_1 v_2}}{g}$$

Last edited by a moderator:

- May 27, 2012

- 47

considering the triangle $ABC$,

$$\begin{align*}

\alpha &=90^\circ - \beta \\

\tan{\alpha} &= \tan (90^\circ -\beta ) \\

\tan{\alpha} &= \cot \beta \\

\frac{gt}{v_1}&=\frac{v_2}{gt}\\

g^2t^2 &=v_1\times v_2\\

\therefore t &= \frac{\sqrt{v_1v_2}}{g} \qquad since \; t>0 \\ \end{align*}

$$

the rest is as same as

- Thread starter
- #5

- Feb 6, 2012

- 20

Thank You!