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Three particles A,B and C are thrown from the top of a tower with the same speed. A is thrown straight up, B is thrown straight down and C is thrown horizontally. They hit the ground with speed V(A),V(B) and V(C) respectively.
Options
(a) V(A)=V(B)=V(C)
(b) V(B)>V(C)>V(A)
(c) V(A)=V(B)>V(C)
(d) V(A)>V(B)=V(C)
Correct Answer:
V(A)=V(B)>V(C)
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - When a beam of light is used to determine the position of an object,
- A particle of mass m is projected with a velocity v making an angle of 45°
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Topics: Motion in Straight Line
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- When a beam of light is used to determine the position of an object,
- A particle of mass m is projected with a velocity v making an angle of 45°
- Two coaxial solenoids are made by winding thin insulated wire over a pipe
- A charged oil drop is suspended in a uniform field at 3×10⁴ V/m so that it neither falls
- The velocity of electromagnetic radiation in a medium of permittivity ε₀
Topics: Motion in Straight Line (93)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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