Question

Two bodies are projected horizontally from the same elevated point in opposite directions with speeds 3 ms and 12 msi respectively. At
the moment when their velocity vectors are mutually perpendicular, the horizontal separation between them is (g = 10 ms 2)

a) 9.8 m
b) 9 m
c) 3.6 m
d) 7.8 m
​

Answer 1

Given:

Two bodies are projected horizontally from the same elevated point in opposite directions with speeds 3 m/s and 12 m/s respectively.

To find:

Separation when velocity vectors are perpendicular ?

Calculation:

Final velocity vector for 1st projectile :

$\vec{v1} =( u1 )\hat{i} + gt ( - \hat{ j})$

Final velocity vector for 2nd projectile :

$\vec{v2} =( u2)( - \hat{i} )+ gt ( - \hat{ j})$

Since these vectors are perpendicular:

$\therefore \: \vec{v1}. \vec{v2} = 0$

$\implies \: (u1)( - u2) + {(gt)}^{2} = 0$

$\implies \: {(gt)}^{2} = (u1)(u2)$

$\implies \: {t}^{2} = \dfrac{(u1)(u2)}{ {g}^{2} }$

$\implies \: {t}^{2} = \dfrac{(3)(12)}{ {(10)}^{2} }$

$\implies \: {t}^{2} = \dfrac{36}{100}$

$\implies \: t = 0.6 \: sec$

Now , separation will be :

$\therefore \: x = v_{rel} \times (t)$

$\implies \: x = (3 + 12) \times 0.6$

$\implies \: x = 15 \times 0.6$

$\implies \: x = 9 \: m$

So, separation will be 9 metres