Q. 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:
EAMCETEAMCET 2005
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.
