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A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v.The two balls meet at t = 18s. What is the value of v?(take g = 10 m/s²)
Options
(a) 75 m/s
(b) 55 m/s
(c) 40 m/s
(d) 60 m/s
Correct Answer:
75 m/s
Explanation:
Clearly distance moved by 1st ball in 18 s = distance moved by 2nd ball in 12 s. Now, distance moved in 18 s by 1st ball 1/2 x 10 x 18²
= 90 x 18 = 1620 m. Distance moved in 12 s by 2nd ball = ut + 1/2 gt²
1620 = 12v + 5 x 144
v = 135 – 60 = 75 ms
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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 boy is trying to start a fire by focusing sunlight on a piece of paper using
- The heat dissipated in a resistance can be obtained by measuring of resistance,
- If a body travels half the distance with velocity v1 and the next half with velocity v2
- The wave described by y = 0.25 sin (10 2πx – 2πt), where x and y are in meters
- The magnifying power of a telescope is 9. When it is adjusted for parallel rays
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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