Thank you for reporting, we will resolve it shortly
Q.
The path of one projectile as seen by an observer on another projectile is a/an:
Motion in a Plane
Solution:
We know that
$x=(u\,\,cos\,\, \theta)t \,\,$ and $\, y=(u \,\,sin \,\,\theta)t- \frac {1}{2} gt^{2}$
Let $x_{2}-x_{1}=(u_{1}\,\,cos\,\, \theta_{1}-u_{2} \,\,cos\,\,\theta_{2})t=X$
$y_{2} -y_{1} =(u_{1}\,\,sin\, \theta_{1})t-\frac{1}{2} gt^{2}$
$-(u_{2}\,\,sin\,\, \theta_2)t +\frac{1}{2} gt^{2}$
$=(u_{1}\,\,sin\,\, \theta_{1} -u_{2}\,\,sin \,\, \theta_{2})t=Y$
$\frac{Y}{X}=\frac{(u_{1}\,\,sin\,\, \theta_{1}- u_{2} \,\,sin\,\ \theta_{2})t}{(u_{1}\,\,cos\,\, \theta_{1}-u_{2}\,\,cos\,\, \theta_{2})t}$
$=\frac{u_{1}\,\,sin\,\, \theta_{1} -u_{2}\,\,sin\,\, \theta_{2}}{u_{1} \,\, cos \theta_{1}-u_{2}\,\,cos\,\, \theta_{2}}$
= constant, m (say)
Y= m X
It is the equation of a straight line passing through the origin.