Question
Prove that the path of one projectile as seen from another projectile is a straight line.
✓
Answer
The coordinates of one projectile as seen from another projectile are: $\text{X}=\text{x}_1-\text{x}_2=(\text{u}_1\cos\theta_1-\text{u}_2\cos\theta_2)\text{t}$ $\text{Y}=\text{y}_1-\text{y}_2$ $=(\text{u}_1\sin\theta_1)\text{t}-\frac{1}{2}\text{gt}^2-(\text{u}_2\sin\theta_2)\text{t}+\frac{1}{2}\text{gt}^2$$=(\text{u}_1\sin\theta_1-\text{u}_2\sin\theta_2)\text{t}$
$\therefore\ \frac{\text{Y}}{\text{X}}=\frac{(\text{u}_1\sin\theta_1-\text{u}_2\sin\theta_2)\text{t}}{(\text{u}_1\cos\theta_1-\text{u}_2\cos\theta_2)\text{t}}$
$=\frac{\text{u}_1\sin\theta_1-\text{u}_2\sin\theta_2}{\text{u}_1\cos\theta_1-\text{u}_2\cos\theta_2}=\text{m}$ (constant)
or, $\text{Y}=\text{mX}$
This equation represents straight line. Hence proved.
$\therefore\ \frac{\text{Y}}{\text{X}}=\frac{(\text{u}_1\sin\theta_1-\text{u}_2\sin\theta_2)\text{t}}{(\text{u}_1\cos\theta_1-\text{u}_2\cos\theta_2)\text{t}}$
$=\frac{\text{u}_1\sin\theta_1-\text{u}_2\sin\theta_2}{\text{u}_1\cos\theta_1-\text{u}_2\cos\theta_2}=\text{m}$ (constant)
or, $\text{Y}=\text{mX}$
This equation represents straight line. Hence proved.
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