| ⇦ | 
| ⇨ |
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:
- A bus is moving with a speed of 10 ms⁻¹ on a straight road
- The velocity of a particle at an instant is 10 ms⁻¹ and after 5 s the velocity
- Bohr’s postulates quantizes
- When a sound wave of wavelength λ is propagating in a medium,
- The maximum number of possible interference maxima for slit-seperation equal to
Topics: Motion in Straight Line
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply