question_answer

A body is projected with velocity \[{{\nu }_{1}}\] at angle \[60{}^\circ \] point A as shown in figure, at the same instant another body is projected vertically upward with velocity \[{{\nu }_{2}}\] from point B (as shown in figure). The value of \[\frac{{{\nu }_{1}}}{{{\nu }_{2}}}\] so that both bodies will collide  

A)  1                                

B)  \[\frac{\sqrt{3}}{2}\]

C)  \[\frac{2}{\sqrt{3}}\]                            

D)  None of these

Correct Answer: C

Solution :

The two bodies will collide, if they reach at a point covering same vertical distance in same time. Let the point of collision be D which is at a height from B, and time of collision be t, then \[\operatorname{h}={{v}_{1}}\,in\,\,60{}^\circ \,\times t-\frac{1}{2}\,g{{t}^{2}},\,\,h={{v}_{2}}t\,-\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,{{v}_{1}}\frac{\sqrt{3}}{2}t-\frac{1}{2}g{{t}^{2}}={{v}^{2}}t-\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\frac{{{v}_{1}}\sqrt{3}}{2}\,\,=\,\,{{v}^{2}}\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{2}{\sqrt{3}}\]