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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
- The electric potential at a point (x,y,z) is given by V = – x²y – xz³ + 4.
- A spring of spring constant 5 x 10³ Nm⁻¹ is streched intially by 5 cm from
- A nucleus with mass number 220 initially at rest emits an α-particle. If the Q-value
- An object placed at 20 cm in front of a concave mirror produces three times
- Two discs of same material and thickness have radii 0.2 m and 0.6 m.
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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