Question

Multiple Correct Answers Type
Suppose two particles 1 and 2 are projected in vertical plane simultaneously.
Their angles of projection are ${30}^{\circ }$ and $\theta$, respectively, with the horizontal. Let they collide after a time t in air. Then

• A. $\theta =si{n}^{-1}\left(4/5\right)$ and they will have same speed just before the collision.
• B. $\theta =si{n}^{-1}\left(4/5\right)$ and they will have different speed just before the collision.
• C. $x<1280\sqrt{3}-960m$
• D. It is possible that the particles collide when both of them are at their highest point.

Answer 1

Answer: (B), (C), (D)

The correct options are
B $x<1280\sqrt{3}-960m$
C $\theta =si{n}^{-1}\left(4/5\right)$ and they will have different speed just before the collision.
D It is possible that the particles collide when both of them are at their highest point.

If they collide, their vertical component of velocities should be same, i.e.
$100\sin \theta =160\sin {30}^{\circ }\Rightarrow \sin \theta =4/5$
Their vertical components will always be same.
Horizontal components:
$160\cos 30=80\sqrt{3}m{s}^{-1}$
and $100\cos \theta =100\times 3/5=60m{s}^{-1}$
They are not same, hence their velocities will not be same at any time. So (b) is correct.
$x=x_1-x_2=160\cos {30}^{\circ }t-100\cos \theta t$
$\Rightarrow x=(80\sqrt{3}-60)t$
Time of flight: $T=\frac{2\times 160\times \sin 30}{g}=16s$ (same for both)
Now $t<T_x\to$ to collide in air
$\Rightarrow \frac{x}{80\sqrt{3}-60}<16\Rightarrow x<1280\sqrt{3}-960$
Since their times of flight are same, they will simultaneously reach their maximum height. So it is possible to collide at the highest point for certain values of $x$.