Question

Suppose two particles 1 and 2 are projected in the vertical plane at the same time. Their angles of projection are ${30}^0$ and $\theta$ respectively with the horizontal and the velocities of projection are $160m/s$ and $100m/s$. The initial separation between the points of projection is $x$. If they collide after time $t$ in air, then

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

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

$100sin\theta =160\sin {30}^{\circ }$
$\Rightarrow sin\theta =\frac{4}{5}$

Their vertical components will always be the same.

Horizontal components are $160\cos 30m/s$ and $100\cos \theta =100\times \frac{3}{5}=60m/s$

They are not same, hence their velocities will not be the same at any time before collision. So (b) is correct.

$x=x_1-x_2=(160cos30-100cos\theta )t$
$\Rightarrow x=(80\sqrt{3}-60)t$


Time of flight for both projectiles $T=\frac{2\times 160sin30}{g}=16s$

For the collsion to happen in air, time of collision
$t<$ time of flight $T$

$\Rightarrow \frac{x}{(80\sqrt{3}-60)}<16$
$\Rightarrow x<1280\sqrt{3}-960$

Since the times of flight are the same, they will simultaneously reach the maximum height. So it is possible for them to collide at the maximum height.