question_answer

Three particles A, B and C are thrown from the top of a tower with the same speed. A is thrown up, B is thrown down and C is horizontally. They hit the ground with speeds \[{{\text{v}}_{\text{A}}}\], \[{{\text{v}}_{\text{B}}}\] and \[{{\text{v}}_{\text{C}}}\] respectively then,          

A) \[{{\text{v}}_{\text{A}}}\text{=}{{\text{v}}_{\text{B}}}\text{=}{{\text{v}}_{\text{C}}}\]       

B)        \[{{\text{v}}_{\text{A}}}\text{=}{{\text{v}}_{\text{B}}}\text{}{{\text{v}}_{\text{C}}}\]

C) \[{{\text{v}}_{\text{A}}}\text{}{{\text{v}}_{\text{C}}}\text{}{{\text{v}}_{\text{B}}}\]

D)        \[{{\text{v}}_{\text{A}}}\text{}{{\text{v}}_{\text{B}}}\text{=}{{\text{v}}_{\text{C}}}\]

Correct Answer: A

Solution :

[a] For A: It goes up with velocity u will it reaches its maximum height (i.e. velocity becomes zero) and comes back to O and attains velocity u..

Using \[{{v}^{2}}={{u}^{2}}+2as\,\,\,\Rightarrow \,\,\,{{v}_{A}}=\sqrt{{{u}^{2}}+2gh}\]

For B, going down with velocity u

\[\Rightarrow {{\text{v}}_{\text{B}}}=\sqrt{{{\text{u}}^{2}}+2\text{gh}}\]

For C, horizontal velocity remains same, i.e. u.

Vertical velocity \[=\sqrt{0+2\text{gh}}=\sqrt{2\text{gh}}\]

The resultant \[{{\text{v}}_{\text{C}}}=\sqrt{{{\text{v}}_{\text{x}}}^{2}+{{\text{v}}_{\text{y}}}^{2}}\text{=}\sqrt{{{\text{u}}^{\text{2}}}\text{+2gh}}.\]

Hence \[{{\text{v}}_{\text{A}}}={{\text{v}}_{\text{B}}}={{\text{v}}_{\text{C}}}\]