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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.
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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
- A particle has initial velocity(3i⃗ +4j⃗)and has acceleration (0.4i⃗ +0.3j⃗)
- In a uniform circular motion, work done in one complete rotation is
- In a series resonant LCR circuit, the voltage across R is 100 volts and R =1kΩ
- When ₃Li⁷ nuclei are bombarded by protons, and the resultant nuclei are ₄Be⁸,
- The ratio of the radii or gyration of a circular disc to that of a circular ring
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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