Question

Suppose two particles 1 and 2 are projected in vertical plane simultaneously.
Their angles of projection are 30${}^o$ and $\theta$ respectively with the horizontal. Suppose 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 $\theta =si{n}^{-1}\left(4/5\right)$ 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.

If they collide, their vertical components of velocities should be the same,

$100sin\theta =160sin{30}^0$

this implies that :

$sin\theta =\frac{4}{5}$

Their vertical components will always be the same. Horizontal components

$160cos{30}^0=80\sqrt{3}m/s$

and $100cos\theta =100\frac{3}{5}=60m/s$

They are not same hence their velocities will not be same at any time ,

$x=x_1-x_2$

$=10cos{30}^{0}t-100cos\theta t$

so $x=(80\sqrt{3}-60)t$

Time of flight, $T=\frac{2\times 160\times sin{30}^0}{g}=16s$

Now t < T - to collide in air,

therefore,

$\frac{x}{80\sqrt{3}-60}<16$ this implies that $x<1280\sqrt{3}-960$

Since their times of flight are same, they will simultaneously reach their maximum height.

So it is possible to collide at highest point for certain values of $x$.