Question

In the figure shown, the two projectiles are fired simultaneously. The minimum distance between
them during their flight is
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Answer 1

The question is incomplete and here is the missing diagram in attachment

minimum distance is 10 m.

Explanation:

Taking the reference point at A and  X- axis along the AB line.

The velocity of second projectile with respect to the first is given by velocity of first projectile-velocity of second projectile.

• The velocity of first projectile(A) is $Va$= 20√3(cos60î+sin60j)
• the velocity of second projectile(B) is $Vb$=20(cos150î+sin150j)

velocity of B with respect to A is $Vab$= $Va-Vb$

$Vab$=20√3î+20j

assume a third particle is moving with this velocity, this implies it is projected with an angle Ф=tan⁻¹$\frac{20}{20\sqrt{3} }$

therefore, it is projected at an angle of 30°.

minimum distance between the two projectiles is when their paths are particular. therefore they are perpendicular when this Ф=30°.

from the below triangle in second figure, minimum distance is $d= 20*sin(30)$

Therefore, $d=10 m$