Question

Two projectiles are projected horizontally from the same point in opposite directions with speed 5 m/s. Find the separation in meter at the instant their velocities become mutually perpendicular (g= = 10 m/s2).​

Answer 1

At the instant shown in the fig., the particles make angle $\alpha$ and $\beta$ each with vertical, such that $\sf{\alpha+\beta=90^o.}$

From the fig. we get,

$\sf{\longrightarrow\tan\alpha=\tan\beta=\dfrac{5}{10t}}$

Since $\alpha$ and $\beta$ are acute angles,

$\longrightarrow \alpha=\beta$

So,

$\sf{\longrightarrow\alpha+\beta=90^o}$

$\sf{\longrightarrow2\alpha=90^o}$

$\sf{\longrightarrow\alpha=\beta=45^o}$

Then,

$\sf{\longrightarrow\tan45^o=\dfrac{5}{10t}}$

$\sf{\longrightarrow1=\dfrac{1}{2t}}$

$\sf{\longrightarrow t=\dfrac{1}{2}\ s}$

The horizontal as well as vertical displacements of both the particles are same, so the separation between them is twice the horizontal displacement of one of them, since they vertically move in same direction but horizontally in opposite direction.

$\sf{\longrightarrow d=2x}$

$\sf{\longrightarrow d=2\cdot5\cdot\dfrac{1}{2}}$

$\sf{\longrightarrow\underline{\underline{d=5\ m}}}$

Answer 2

Given:

Two projectiles are projected horizontally from the same point in opposite directions with speed 5 m/s.

To find:

Find the separation in meter at the instant their velocities become mutually perpendicular

Calculation:

Let us assume that the factors of the projectile become mutually perpendicular after time t :

Let the final Velocity vectors be v1 and v2:

For 1st projectile :

$\therefore \: \vec{v}_{1} = u \: \hat{i} + gt \: \hat{j}$

For 2nd Projectile (which is moving in opposite directions) :

$\therefore \: \vec{v}_{2} = u \: ( - \hat{i} )+ gt \: \hat{j}$

$\implies \: \vec{v}_{2} = - u \: \hat{i} + gt \: \hat{j}$

Now , since these vectors ars perpendicular :

$\therefore \: \vec{v}_{1} \: . \: \vec{v}_{2} = 0$

$\implies\: - {u}^{2} + {g}^{2} {t}^{2} = 0$

$\implies\: {g}^{2} {t}^{2} = {u}^{2}$

$\implies\: {t}^{2} = {( \dfrac{u}{g} )}^{2}$

$\implies\: t= \dfrac{u}{g}$

$\implies\: t= \dfrac{5}{10}$

$\implies\: t= 0.5 \: sec$

Now , the separation between the projectile at that time :

$\therefore \: d = 2 \times (u \times t)$

$\implies \: d = 2 \times 5 \times 0.5$

$\implies \: d =5 \: metres$

So , separation will be 5 metres.