Question

@yrelhan4

Answer 1

\[\large x_1=(u_1\cos \theta_1)t_1~,y_1=(u_1 \sin \theta_1)t-\frac{1}{2}g t^2\]
\[\large x_2=(u_2\cos \theta_2)t_2~\And~y_2=(u_2 \sin \theta_2)t-\frac{1}{2}g t^2\]
The position of one projectile w.r.t another projectile is:
\[\large x=x_1-x\]
\[\large y=y_1-y_2\]
\[\frac{y}{x}=\frac{u_1 \sin \theta_1-u_2 \sin \theta_2}{u_1 \cos \theta_1-u_2 \cos \theta_2}=constant\]
\[\LARGE y ~\alpha ~x\]
Equation of straight line
Thus,Motion of a projectile as observed from another projectile is a straight line

Answer 2

:O

Answer 3

:)

Answer 4

@yrelhan4 bc tujhe kisne medal aur kyun dia -_-

Answer 5

I'm the boss. \m/ :D

Answer 6

._.