Question

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

Answer 1

Answer: (C)

vertical straight line

Answer 2

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.