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NEET 2010 (AIPMT, Prelims) · Motion Under Gravity

Motion Under Gravity question from NEET 2010 (AIPMT, Prelims)

Q50NEET 2010 · AIPMT · PRELIMSMedium

Motion Under Gravity

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 = 18 s. What is the value of v?

(Take g = 10 m/s2)

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Answer: (a)

  1. First ball at t=18 st = 18\text{ s} has fallen 12(10)(18)2=1620 m\frac{1}{2}(10)(18)^2 = 1620\text{ m}.
  2. Second ball moves for 18−6=12 s18 - 6 = 12\text{ s}: 12v+12(10)(12)2=12v+72012v + \frac{1}{2}(10)(12)^2 = 12v + 720
  3. They meet: 12v+720=1620⇒v=75 m/s12v + 720 = 1620 \Rightarrow v = 75\text{ m/s}
  4. So the answer is (a) 75 m/s.
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