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Two particles are projected simultaneously in a vertical plane from the same point. These particles have different velocities and different angles with the horizontal. The path seen by each other is:
(A) Parabola
(B) Hyperbola
(C) Elliptical
(D) Straight line

Last updated date: 13th Jun 2024
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Hint: Resolve the velocities of the particles in x and y coordinates.
Evaluate the relative velocity. From the expression of relative velocity, predict the curve.

Complete step by step solution:
Let $\overrightarrow {{u_1}} \& \overrightarrow {{u_2}} $ be the initial velocities of particles and let the particles make angles of ${\theta _1}\& {\theta _2}$ with the horizontal respectively.
After time t, their velocities will be $\overrightarrow {{v_1}} = ({u_1}\cos {\theta _1})\widehat i + ({u_1}\cos {\theta _1} - gt)\widehat j$
And similarly $\,\overrightarrow {{v_2}} = ({u_2}\cos {\theta _2})\widehat i + ({u_2}\cos {\theta _2} - gt)\widehat j$
Relative velocity $\overrightarrow {{v_{12}}} = ({u_1}\cos {\theta _1} - {u_2}\cos {\theta _2})\widehat i + ({u_1}\sin {\theta _1} - gt - {u_2}\sin {\theta _2} + gt)\widehat j$
$ \Rightarrow \overrightarrow {{v_{12}}} = ({u_1}\cos {\theta _1} - {u_2}\cos {\theta _2})\widehat i + ({u_1}\sin {\theta _1} - {u_2}\sin {\theta _2})\widehat j$
It is clear that relative velocity is independent of time and any other possible variables, this means that relative velocity of a projectile is constant.
Therefore, the path seen by a particle is a straight line making an angle with the horizontal.

Therefore, option (D) is correct.

Note: Curves of common equations are as follows:
Linear – straight line
Quadratic – parabola
In questions like these, try to express the equation in terms of power of t. The power of t decides the shape of the curve.