Answer to Question #92289 in Mechanics | Relativity for Sridhar

Question

Two stones are projected from the top of a tower in opposite directions with the same velocity V at angles 30°,60° to the horizontal. What is the relative velocity of one stone with respect to other stone?

Answer 1

Considering stone 1 going in positive $x-axis$ and stone 2 going in negative direction of $x-axis$

The vertical velocity of both stones is in same direction(+$ve$ ).

Stone 1:

Angle with horizontal=$30\degree$

Horizontal velocity($h_1$ )=$Vcos\theta=Vcos 30\degree =+\frac{\sqrt{3}}{2}V \hat{i}$

Vertical velocity($v_1$ )$=Vsin\theta=Vsin 30\degree=+\frac{1}{2}V\hat{j}$

Net velocity of stone 1=$V_1=+\frac{\sqrt{3}}{2}V\hat{i}+\frac{1}{2}\hat{j}$

Stone 2:

Angle with horizontal=$60\degree$

Horizontal velocity=$Vcos\theta=Vcos 60\degree=-\frac{1}{2}V\hat{i}$

Vertical velocity$=Vsinθ=Vsin60°=+\frac{\sqrt{3}}{2}V\hat{j} ​$

Net velocity of stone 2=$V_2=-\frac{V}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$

Relative velocity =$V_{12}=V_1-V_2$

=$(\frac{\sqrt{3}}{2}V+\frac{V}{2})\hat{i}+(\frac{V}{2}-\frac{\sqrt{3}V}{2})\hat{j}$