# Homework Help: Am I correct?

1. Dec 17, 2005

### vaishakh

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.
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.

2. Dec 17, 2005

### neutrino

It for the angle between the vectors. What happens to the dot product of two vectors when the angle between them is 90°?

3. Dec 17, 2005

### mukundpa

if the two velocity vectors are perpendicular and one is making angle theeta with downward vertical the other will make an angle 90 - theeta with downwaed vertical. hence if for first tan theeta = 3/gt for the other cot theeta = 4/gt gives g^2*t^2 = 12 may get time easily.