See this question. Two particles are moving in a uniform gravitational field with an acceleration g. at the initial moment, the particles were located at one point and moved with velocities v1 = 3m/sec and v2 = 4m/sec horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular to each other.(adsbygoogle = window.adsbygoogle || []).push({});

I wrote the velocity of the particles as a function of time. Let i be a unit vector horizontally j be a unit vector vertically.

Thus v1(t) = 3ti - 4.9t^2j and v2(t) = -4ti - 4.9t^2j.

Then I expressed theta1 and theta2, the angles made by the velocity vector of the particle with the vertical. There theta1(t) = tan^-1(4.9t/3) and theta2 = tan^-1(4.9t/4). Since the angle became 90deg, we can write the equation,

Tan^-1(4.9t/3) + tan^-1(4.9t/4) = pi/2.

Therefore 4.9t/3 = 4/4.9t. Therefore t^2 = 2^1/2 approx.

The rest of the work is normal and can be solved just using projectile equations. But I want to know whether my approach to the problem is good or whether there is another better approach. The doubt is because this is the first time I am using such a method and it is totally new to me. I hope the method is correct.

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# Homework Help: Am I correct?

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