question_answer

A body is projected horizontally from the top of a tower with an initial velocity of 18 m/s. It hits the ground at an angle of\[45{}^\circ \]. What is the vertical component of velocity when the body strikes the ground?

A) \[9\,m/s\]                      

B) \[9\sqrt{2}\,m\]

C) \[18\,m/s\]                     

D) \[18\sqrt{2}\,m/s\]

Correct Answer: C

Solution :

\[\because \,\,\,\,\tan \alpha \,\,=\,\,\frac{{{V}_{y}}}{{{V}_{x}}},\,\,\,\,\,\,\,\,\,\,\,\,1=\frac{{{V}_{y}}}{18}\] \[\therefore \,\,\,\text{ }{{V}_{y}}\,\,=\,\,18\text{ }m/s\]