question_answer

A man standing on the roof of a house of height h throws one particle vertically downwards and another particle horizontally with the same velocity u. The ratio of their velocities when they reach the earth?s surface will be

A) \[\sqrt{2gh+{{u}^{2}}}:u\]

B)        \[1:2\]

C) \[1:1\]                           

D) \[\sqrt{2gh+{{u}^{2}}}:\,\,\sqrt{2\,gh}\]

Correct Answer: C

Solution :

When particle thrown in vertically downward direction with velocity u then final velocity at the ground level. \[{{\operatorname{v}}^{2}}={{u}^{2}}+2gh\,\,\,\,\,\,\therefore \,\,v\,\,=\,\,\sqrt{{{u}^{2}}+2gh}\] Another particle is thrown horizontally with same velocity then velocity of particle at the surface of earth. \[Horizontal\,\,component\,\,of\,\,velocity\,\,{{v}_{x}}=u\] \[\therefore \,\,\,\,Resultant\,\,velocity,\,\,\,\,\nu \,\,={{u}^{2}}\,\,+\,\,2gh\] For both the particles, final velocities when they reach the earth?s surface are equal.