Kinematics Question 644

Question: 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,

Options:

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}}} $

Show Answer

Answer:

Correct Answer: A

Solution:

[a] For A: It goes up with velocity u will it reaches it’s 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{{{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}}} $



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