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}}} $
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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}}} $
BETA
