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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
Related Questions: - A simple pendulum has a time period T₁ when on the earth’s surface and T₂ When taken
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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 simple pendulum has a time period T₁ when on the earth’s surface and T₂ When taken
- Two equal and opposite charges of masses m₁ and m₂ are accelerated
- If the half-life of a material is 10 yr, then in what time, it becomes 1/4th part
- Under the action of a force F=cx, the position of a body changes from 0 to x.The work done
- A silver wire of radius 0.1 cm carries a current of 2A. If the charge density in silver
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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