Question

Two particles A and B are projected simultaneously in the direction shown in the figure with velocities $v_A=20m/s$ and $v_B=10m/s$ respectively. They collide in air after $\frac{1}{2}$ sec. Then

• A. the angle $\theta$ is ${60}^{\circ }$
• B. the distance $x$ is $5\sqrt{3}m$
• C. the angle $\theta$ is ${30}^{\circ }$
• D. the distance $x$ is $15\sqrt{3}m$

Answer 1

The correct option is A the angle $\theta$ is ${60}^{\circ }$

Time interval$=\frac{1}{2}$

We know that, at maximum velocity becomes zero

$V_{AY}=V\sin \theta \to \left(1\right)V_{BY}=V\sin 90°\to \left(2\right)$

Now, we can calculate time at which velocity becomes zero.

$v=u+at$(Newton's equation)

$0=V_A\sin \theta -g{t}_A{t}_A=\frac{V_A\sin \theta }{g}0=V_B\sin 90°-g{t}_B{t}_B=\frac{V_B}{g}$

Now, we can calculate maximum height

$s=ut+\frac{1}{2}a{t}^2{v}^2+u^2+2as0=(V_A\sin \theta {)}^2-2g\times S_A$

and $0=(V_B\sin 90°{)}^2-2g{S}_B{S}_B=\frac{V_A^2\sin 90°}{2g}$

Now, $S_B=\frac{V_B^2}{2g}$

$t_A-t_B=\frac{1}{2}\frac{V_A\sin \theta }{g}-\frac{V_B}{g}=\frac{1}{2}\frac{1}{g}(V_A\sin \theta -V_B)=\frac{1}{2}10=(20\sin \theta -10)\times 25=20{\sin }^2\theta -1020{\sin }^2\theta =15{\sin }^2\theta =\frac{15}{20}\sin \theta =\frac{\sqrt{3}}{2}\theta =60°$