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)

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.