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Two particles are projected simultaneously in the same vertical plane, from the same point on ground, but with same speeds but at different angles $( < 90^o )$ to the horizontal. The path followed by one, as seen by the other, is
a vertical straight line
a straight line making a constant angle with the horizontal
a parabola
a hyperbola
Solution
Let $\vec{u}_{1}$ and $\vec{u}_{2}$ be the initial velocities of the two particles and $\theta_{1}$ and $\theta_{2}$ be their angles of projection with the horizontal.
The velocities of the two particles after time $t$ are,
$\vec{v}_{1}=\left(u_{1} \cos \theta_{1}\right) \hat{i}+\left(u_{1} \sin \theta_{1}-\mathrm{gt}\right) \hat{j}$
$\vec{v}_{2}=\left(u_{2} \cos \theta_{2}\right) \hat{i}+\left(u_{2} \sin \theta_{2}-\mathrm{gt}\right) \hat{j}$
$\vec{v}_{12}=\vec{v}_{1}-\vec{v}_{2}=\left(u_{1} \cos \theta-u_{2} \cos \theta_{2}\right) \hat{i}+\left(u_{1} \sin \theta_{1}-\mathrm{gt}-u_{2} \sin \theta_{2}+\mathrm{gt}\right) \hat{j}$
which is a constant So the path followed by one, as seen by the other is straight line, making a constant angle with the horizontal.
