Question

Show that the motion of one projectile as seen from another projectile will always be a straight line motion. ........... with full explanation..........​

Answer 1

Suppose Ua and Ub are the initial velocities of two projectiles and ‘’x’’ and ‘’y’’ are consider as direction. So components of relative velocities in both directions will be,

$\mathrm{U}_{\mathrm{abx}}=\mathrm{U}_{\mathrm{ax}}-\mathrm{U}_{\mathrm{bx}} \text { and } \mathrm{U}_{\mathrm{aby}}=\mathrm{U}_{\mathrm{ay}}-\mathrm{Uby}$

Therefore relative velocities between these two projectiles will be,

$\mathrm{U}_{\mathrm{ab}}=\left(\mathrm{U}_{\mathrm{ax}}-\mathrm{U}_{\mathrm{bx}}\right) \mathrm{i}+\left(\mathrm{U}_{\mathrm{ay}}-\mathrm{U}_{\mathrm{by}}\right) \mathrm{j}$

In this case resultant velocity will be constant and it will make an angle with horizontal axis. Therefore motion of one projectile when viewed from other projectile will be a straight line motion.  

Answer 2

Explanation:

If $\sf (x_A,y_A)$ is the position of projectile A after time $\sf t$ projected with a velocity $\sf v_0^A$,then

• $\sf x_A=v_{x0}^At$
• $\sf y_A=v_{y0}^At - \dfrac{1}{2}gt^2$

Similarly, for projectile B with projection $\sf v_0^B$,

• $\sf x_B=v_{x0}^Bt$
• $\sf y_B=v_{y0}^Bt - \dfrac{1}{2}gt^2$

$\sf (x_B-x_A)=(v_{x0}^B-v_{x0}^A)$

$\sf (y_B-y_A)=(v_{y0}^B-v_{y0}^A)$

$\sf \dfrac{y_B-y_A}{x_B-xA} = \dfrac{v_{y0}^B-v_{y0}^A}{v_{x0}^B-v_{x0}^A} = m$

Let $\sf (x,y)$ represent position of a projectile B relative to projectile A. Then $\sf x_B-x_A=x$ and $\sf y_B-y_A=y$

Thus, $\sf \dfrac{y}{x}=m$ or $\sf y= mx$, which is equation of a straight line.