Question

Suppose two particle, $1$ and $2$ are projected in vertical plane simultaneously with the velocities of $160\text{m/s}$ and $100\text{m/s}$ respectively as shown in the figure.


Their angles of projection are ${30}^{\circ }$ and $\theta$, respectively, with the horizontal. If they collide after time $t$ in air, then which of the following is incorrect?

• A. $\theta ={53}^{\circ }$ and they will have same speeds just before the collision.
• B. $\theta ={53}^{\circ }$ and they will have different speeds just before the collision.
• C. $x<(1280\sqrt{3}-960)\text{m}$
• D. It is possible that the particles collide when both of them are at their highest point.

Answer 1

The correct option is A $\theta ={53}^{\circ }$ and they will have same speeds just before the collision.

If the particles have to collide, the vertical components of their velocities should be same.
i.e. $100\sin \theta =160\sin {30}^{\circ }\Rightarrow \sin \theta =\frac{4}{5}$
$\Rightarrow \theta ={53}^{\circ }$
Horizontal component of velocity for particle 1:
$=160\cos {30}^{\circ }=80\sqrt{3}\text{m/s}$
and horizontal component of velocity for particle 2:
$=100\cos \theta =100\times \frac{3}{5}=60\text{m/s}$
Since horizontal components are not same and vertical components are same, their final velocities will be different at any time. So (b) is correct.


$x=x_3-x_2=160t\cos {30}^{\circ }-100t\cos \theta$
$\Rightarrow x=(80\sqrt{3}-60)t$
Time of flight: $T=\frac{2\times 160\times \sin {30}^{\circ }}{g}=16\text{s}$
Now $t<T$ to collide in air
$\Rightarrow \frac{x}{80\sqrt{3}-60}<16\Rightarrow x<1280\sqrt{3}-960$
Since their times of flight are the same, they will simultaneously reach their maximum height. So, it is possible to collide at highest point for certain values of $x$.
i.e (c) and (d) are also correct. Hence, only option (a) is incorrect.

Alternative solution:
Let $y_1$ and $y_2$ be the vertical distances covered by particle 1 and 2 respectively.
$\tan \alpha =\frac{y_2-y_1}{x_2-x_3}=\left|\frac{(u_{12}{)}_y}{(u_{12}{)}_x}\right|$
Since $y_1=y_2\Rightarrow (u_{12}{)}_y=0\Rightarrow u_{1y}=u_{2y}$
$\Rightarrow 160\sin {30}^{\circ }=100\sin \theta \Rightarrow \sin \theta =\frac{4}{5}\Rightarrow \theta ={53}^{\circ }$
In the vertical direction: $(u_{12}{)}_y=0;(g_{12})=0$ and $(S_{12}{)}_y=0$
$\Rightarrow (v_{12}{)}_y=0$
Hence, they will have same final velocities in vertical direction.
But in horizontal direction, $u_{1x}\ne u_{2x}$
b) Hence, their net velocities will be different.
c) For them to collide in air
$t<T$ (time of flight of any one)
$\frac{x_3-x_2}{(v_{12}{)}_x}<\frac{2\times 100\times \frac{4}{5}}{g}$
$\Rightarrow \frac{x}{|160\cos {30}^{\circ }-100\cos \theta |}<16$
$\Rightarrow x<16[80\sqrt{3}-60]$
Since their time of flight are same, they will reach their maximum height simultaneously. So, it is possible to collide at highest point for certain values of $x$.
Hence, only option (a) is incorrect.