question_answer

Two identical particles are projected horizontally in opposite directions with a speed of \[5\text{ m}{{\text{s}}^{-1}}\] each from the top of a tall tower as shown. Assuming \[\text{g = 10 m}{{\text{s}}^{-2}}\], the distance between them at the moment when their velocity vectors become mutually perpendicular is

A) 2.5 m               

B) 5 m

C) 10 m                            

D) 20 m

Correct Answer: B

Solution :

[b] At a time t when velocity vector become mutually perpendicular \[v\text{ }cos\text{ }45{}^\circ =5\] horizontal component

\[v=\frac{5}{cos\text{ }45{}^\circ }=5\sqrt{2}\text{ m/s }\]

Vertically,

\[v\,\,\sin 45{}^\circ =gt\]

\[\Rightarrow t=\frac{v\,\,\sin 45{}^\circ }{g}=\frac{5\sqrt{2}\times 1/\sqrt{2}}{10}=\frac{5}{10}=\frac{1}{2}\]

So, \[OA=OB=v\cos 45{}^\circ \times t=5\times \frac{1}{2}=2.5\]

\[\Rightarrow \,\,\,AB=2.5\times 2=5\text{ m}\]