question_answer

Two balls are projected simultaneously in the same vertical plane from the same point with velocities \[{{V}_{1}}\]and \[{{V}_{2}}\]with angles \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] respectively with the horizontal. If , the path of one ball as seen from the position of other ball is:

A)  parabola

B)  horizontal straight line

C)  vertical straight line

D)   straight line making \[45{}^\circ \] with the vertical

Correct Answer: C

Solution :

                 For the ball projected with velocity \[{{V}_{1}}\] at an angle \[{{\theta }_{1}}\] with horizontal line, the horizontal distance covered after t time.                 \[{{x}_{1}}={{V}_{1}}\cos {{\theta }_{1}}t\] Similarly, for second ball throw with velocity \[{{V}_{2}}\]at an angle \[{{\theta }_{2}}\] with horizontal, horizontal distance covered after time t. \[{{x}_{2}}={{V}_{2}}\cos {{\theta }_{2}}t\] The vertical distances covered are \[{{y}_{1}}={{V}_{1}}\sin {{\theta }_{1}}t-\frac{1}{2}g{{t}^{2}}\] and        \[{{y}_{2}}={{V}_{2}}\sin {{\theta }_{2}}t-\frac{1}{2}g{{t}^{2}}\] \[\therefore \]   \[{{x}_{2}}-{{x}_{1}}=({{V}_{2}}\cos {{\theta }_{2}}-{{V}_{1}}\cos {{\theta }_{1}})t\] and \[{{y}_{2}}-{{y}_{1}}=({{V}_{2}}\sin {{\theta }_{2}}-{{V}_{1}}\sin {{\theta }_{1}})t\] \[\therefore \] \[\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{{{V}_{2}}\sin {{\theta }_{2}}-{{V}_{1}}\sin {{\theta }_{1}}}{{{V}_{2}}\cos {{\theta }_{2}}-{{V}_{1}}\cos {{\theta }_{1}}}\] but \[{{V}_{1}}\cos {{\theta }_{1}}={{V}_{2}}\cos {{\theta }_{2}}\] \[\therefore \] \[\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{{{V}_{2}}\sin {{\theta }_{2}}-{{V}_{1}}\sin {{\theta }_{1}}}{0}=\infty \] \[\Rightarrow \]               \[{{x}_{2}}-{{x}_{1}}=0\] and        \[{{y}_{2}}-{{y}_{1}}=\infty \] This means line joining the position of particles after time t will be a straight line and parallel to the y-axis.