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NEET 1992 (AIPMT) · Motion Under Gravity

Motion Under Gravity question from NEET 1992 (AIPMT)

Q62NEET 1992 · AIPMTMedium

Motion Under Gravity

A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is (g=10 m/s2g = 10\text{ m/s}^2)

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Answer: (b)

  1. Total time TT: 12gT2−12g(T−2)2=40\frac{1}{2}gT^2 - \frac{1}{2}g(T-2)^2 = 40
  2. 5[T2−(T−2)2]=5(4T−4)=40⇒T=35[T^2 - (T-2)^2] = 5(4T - 4) = 40 \Rightarrow T = 3 s
  3. H=12×10×32=45H = \frac{1}{2} \times 10 \times 3^2 = 45 m
  4. So the answer is (b) 45 m.
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