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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 Zenor diode has a contact potential of 1 V in the absence of biasing. It undergoes
- On a frictionless surface, a block of mass M moving at speed v collides elastically
- A particle in simple harmonic motion is described by the displacement function
- Element ²³⁴U has 92 protons and 234 nucleons in its nucleus. It decays by
- A black body at a temperature of 227⁰C radiates heat at a rate of 20 cal m⁻²s⁻¹.
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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