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There are two interesting cases. What if horizontal component of velocities of the two projectiles are same? In this case, relative velocity of projectiles in horizontal direction is zero. Also, it is imperative that there is no change in the initial separation between two projectiles in x-direction. Mathematically,
$${v}_{ABx}={u}_{Ax}-{u}_{Bx}=0$$
There is no relative motion between two projectiles in horizontal direction. They may, however, move with respect to each other in y-direction. The relative velocity in y-direction is given by :
$${v}_{ABy}={u}_{Ay}-{u}_{By}$$
Since relative velocity in x-direction is zero, the relative velocity in y-direction is also the net relative velocity between two projectiles.
In the second case, components of velocities in y-direction are equal. In this case, there is no relative velocity in y-direction. The projectiles may, however, have relative velocity in x-direction. As such the relative velocity in y-direction is also the net relative velocity between two projectiles.
Two projectiles are projected simultaneously at same speeds, but with different angles of projections from a common point in a given vertical plane. The path of one projectile, as viewed from other projectile is :
(a) a straight line parallel to horizontal
(b) a straight line parallel to vertical
(c) a circle
(d) None of above
The component relative velocities in horizontal and vertical directions are, defined in terms of initial velocities, which are constant for the given pair of projectiles. Therefore, the relative velocities of two projectiles in horizontal and vertical directions are constants. Let " ${u}_{A}$ ” and “ ${u}_{B}$ ” be the initial velocities of two projectiles, then component relative velocities in "x" and "y" directions are :
$${v}_{ABx}={u}_{Ax}-{u}_{Bx}$$ $${v}_{ABy}={u}_{Ay}-{u}_{By}$$
The resultant relative velocity is :
$${\mathbf{u}}_{AB}=\left({u}_{Ax}-{u}_{Bx}\right)\mathbf{i}+\left({u}_{Ay}-{u}_{By}\right)\mathbf{j}$$
As given in the question, the initial speeds of the projectiles are same, but angles of projections are different. Since sine and cosine of two different angles are different, it follows that component velocities of two projectiles are different in either direction. This is ensured as speeds of two projectiles are same. It implies that components (horizontal or vertical) of the relative velocity are non-zero and finite constant. The resultant relative velocity is, thus, constant, making an angle " $\theta $ " with horizontal (x-axis) such that :
$$\mathrm{tan}\theta =\frac{\left({u}_{Ay}-{u}_{By}\right)}{\left({u}_{Ax}-{u}_{Bx}\right)}$$
Thus, one projectile (B) sees other projectile (A) moving in a straight line with constant velocity, which makes a constant angle with the horizontal. Hence, option (d) is correct.
Two projectiles are projected simultaneously at different velocities from a common point in a given vertical plane. If the components of initial velocities of two projectiles in horizontal direction are equal, then the path of one projectile as viewed from other projectile is :
(a) a straight line parallel to horizontal
(b) a straight line parallel to vertical
(c) a parabola
(d) None of above
The component relative velocities in horizontal and vertical directions are, defined in terms of initial velocities, which are constant for the given projectiles. Let " ${u}_{A}$ ” and “ ${u}_{B}$ ” be the initial velocities of two projectiles, then component relative velocities in "x" and "y" directions are :
$${v}_{ABx}={u}_{Ax}-{u}_{Bx}$$ $${v}_{ABy}={u}_{Ay}-{u}_{By}$$
According to the question, components of initial velocities of two projectiles in horizontal direction are equal :
$${u}_{Ax}-{u}_{Bx}=0$$
It is given that velocities of projections are different. As horizontal components of velocities in horizontal directions are equal, the components of velocities in vertical directions are different. As such, above expression evaluates to a constant vector. Thus, one projectile sees other projectile moving in a straight line parallel to vertical (i.e. direction of unit vector j ).
Hence, option (b) is correct.
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