Question

Two particles are projected simultaneously from the same point with the same speed but with different angles of projection $\alpha$ and $\beta (\alpha >\beta )$. Then,

• A. The line joining the positions of the particle at any subsequent time makes a constant angle $\frac{\pi }{2}+\left(\frac{\alpha -\beta }{2}\right)$ with the horizontal.
• B. The line joining the positions of the particle at any subsequent time makes a constant angle $\frac{\pi }{2}+\left(\frac{\alpha +\beta }{2}\right)$with the horizontal.
• C. The magnitude of the relative velocity of the first particle with respect to the second is $2usin\left(\frac{\alpha +\beta }{2}\right)$
• D. The magnitude of the relative velocity of the first particle with respect to the second is $2usin\left(\frac{\alpha -\beta }{2}\right)$

Answer 1

Answer: (B), (C)

The correct options are
B The line joining the positions of the particle at any subsequent time makes a constant angle $\frac{\pi }{2}+\left(\frac{\alpha +\beta }{2}\right)$with the horizontal.
C The magnitude of the relative velocity of the first particle with respect to the second is $2usin\left(\frac{\alpha +\beta }{2}\right)$

For the first particle $x_1=utcos\alpha ,y_1=utsin\alpha -\frac{1}{2}g{t}^2{v}_{1x}=ucos\alpha ;v_{iy}=usin\alpha -gt;\text{For the second particle,}{x}_2=utsin\beta ,y_2=utsin\beta -\frac{1}{2}g{t}^2{v}_{2x}=ucos\beta ;v_{2y}=usin\beta -gt$

Angle made by the line joining the positions of two particles is$tan\theta =\frac{y_2-y_1}{x_2-x_1}=\frac{u(sin\alpha -sin\beta )}{u(cos\alpha -cos\beta )}=-cot\left(\frac{\alpha +\beta }{2}\right)\Rightarrow \theta =\frac{\pi }{2}+\left(\frac{\alpha +\beta }{2}\right)v_{12x}=u(cos\alpha -cos\beta )and{v}_{12y}=u(sin\alpha -sin\beta ).\text{Hence}{v}_{12}=u\sqrt{(cos\alpha -cos\beta {)}^2+(sin\alpha -sin\beta {)}^2}=\sqrt{2-2cos\alpha cos\beta +2sin\alpha sin\beta }=2usin\left(\frac{\alpha +\beta }{2}\right)$