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Question: Two particles are projected simultaneously in a vertical plane from the same point. These particles ...

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

Explanation

Solution

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 u1&u2\overrightarrow {{u_1}} \& \overrightarrow {{u_2}} be the initial velocities of particles and let the particles make angles of θ1&θ2{\theta _1}\& {\theta _2} with the horizontal respectively.
After time t, their velocities will be v1=(u1cosθ1)i^+(u1cosθ1gt)j^\overrightarrow {{v_1}} = ({u_1}\cos {\theta _1})\widehat i + ({u_1}\cos {\theta _1} - gt)\widehat j
And similarly v2=(u2cosθ2)i^+(u2cosθ2gt)j^\,\overrightarrow {{v_2}} = ({u_2}\cos {\theta _2})\widehat i + ({u_2}\cos {\theta _2} - gt)\widehat j
Relative velocity v12=(u1cosθ1u2cosθ2)i^+(u1sinθ1gtu2sinθ2+gt)j^\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
v12=(u1cosθ1u2cosθ2)i^+(u1sinθ1u2sinθ2)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.