Question

The path of one projectile as seen by an observer on another projectile is a/an :

Moderate

Solution

We know that;
$\begin{array}{l} x = (ucos θ) t and y = (usin θ) t - \frac{1}{2} {gt}^{2} \\ Let x_{2} - x_{1} = (u_{1} \cos θ_{1} - u_{2} \cos θ_{2}) t = x \\ y_{2} - y_{1} = (u_{1} \sin θ_{1}) t - \frac{1}{2} {gt}^{2} - (u_{2} \sin θ_{2}) t + \frac{1}{2} {gt}^{2} \\ = (u_{1} \sin θ_{1} - u_{2} \sin θ_{2}) t = Y \end{array}$
$\begin{array}{l} ∴ \frac{Y}{X} = \frac{(u_{1} \sin θ_{1} - u_{2} \sin θ_{2}) t}{(u_{1} \cos θ_{1} - u_{2} \cos θ_{2}) t} \\ = \frac{u_{1} \sin θ_{1} - u_{2} \sin θ_{2}}{u_{1} \cos θ_{1} - u_{2} \cos θ_{2}} \\ = constant, m (say) \\ ∴ Y = mX \end{array}$
It is the equation of straight line passing through the origin.

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