Physics Two Dimensional Motion Question 192
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 $ {v} _A $ , $ {v} _B $ and $ {v} _C $ respectively then,
Options:
A) $ {v} _A\text{=}{v} _B\text{=}{v} _C $
B)$ {v} _A\text{=}{v} _B\text{>}{v} _C $
C) $ {v} _A\text{>}{v} _C\text{>}{v} _B $
D)$ {v} _A\text{>}{v} _B\text{=}{v} _C $
Show Answer
Answer:
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 v _{B}=\sqrt{u^2}+2gh $
For C, horizontal velocity remains same, i.e. u.
Vertical velocity $ =\sqrt{0+2\text{gh}}=\sqrt{2\text{gh}} $
The resultant $ {v} _C=\sqrt{{v _x}^{2}+{v _y}^{2}}\text{=}\sqrt{{u^2}\text{+2gh}}. $
Hence $ {v} _A={v} _B={v} _C $
BETA
