question_answer

A projectile of mass m is thrown with a velocity v making an angle \[60{}^\circ \] with the horizontal. Neglecting air resistance, the change in velocity from the departure A to its arrival at B, along the vertical direction is

A) 2v                    

B)        \[\sqrt{3}\text{v}\]

C) v                     

D)        \[\frac{\text{v}}{\sqrt{3}}\]

Correct Answer: B

Solution :

[b] As the figure drawn above shows that at points A and B the vertical component of velocity is \[v\,\,\sin \,60{}^\circ \] but their directions are opposite. Hence, change in velocity. \[\Delta v=v\text{ }\sin 60{}^\circ -\left( -\,v\sin 60{}^\circ  \right)=2v\,\,\sin 60{}^\circ =\sqrt{3}v\]