Suppose two particles 1 and 2 are projected in vertical plane simultaneously.

Their angles of projection are 30^o and $\mathrm{\theta }$, respectively, with the horizontal. Let they collide after a time t in air. Then 

If they collide, their vertical component of velocities should be same, i.e., 

$\begin{array}{l}100\sin \mathrm{\theta }=160\sin {30}^{\circ }\\ \Rightarrow \sin \mathrm{\theta }=4/5\end{array}$

Their vertical components will always be same. Horizontal components: 

$\begin{array}{l} 160\cos 30=80\sqrt{3}{\mathrm{ms}}^{-1}\\ \mathrm{and} 100\cos \mathrm{\theta }=100\times 3/5=60{\mathrm{ms}}^{-1}\end{array}$

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

$\begin{array}{l} \mathrm{x}={\mathrm{x}}_1-{\mathrm{x}}_2=160\cos {30}^{\circ }\mathrm{t}-100\cos \mathrm{\theta t}\\ \Rightarrow \mathrm{x}=(80\sqrt{3}-60)\mathrm{t}\end{array}$

$\text{ Time of flight: }\mathrm{T}=\frac{2\times 160\times \sin 30}{\mathrm{g}}=16\mathrm{s}\text{ (same for both) }$

Now $\mathrm{t}<{\mathrm{T}}_{\mathrm{x}}\to$ to collide in air 

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

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.