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JEE Main 2019 · Physics · Kinematics and Laws of Motion · easy

A stone is dropped from the top of a tower 80 m high. At the same instant a second stone is thrown vertically upward from the foot of the tower with…

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A stone is dropped from the top of a tower 80 m80\ \mathrm{m} high. At the same instant a second stone is thrown vertically upward from the foot of the tower with speed 40 m s−140\ \mathrm{m\,s^{-1}}. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}. The height above the ground at which the stones pass each other is
  1. A. 60 m60\ \mathrm{m}
  2. B. 40 m40\ \mathrm{m}
  3. C. 20 m20\ \mathrm{m}
  4. D. 50 m50\ \mathrm{m}
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Answer: (A) 60 m60\ \mathrm{m}

Both stones have the same acceleration gg downward, so relative to each other they move uniformly: the relative speed is the constant 40 m s−140\ \mathrm{m\,s^{-1}} and the initial separation is 80 m80\ \mathrm{m}. They meet when t=80/40=2 st = 80/40 = 2\ \mathrm{s}. In this time the dropped stone falls 12gt2=12(10)(4)=20 m\tfrac12 g t^2 = \tfrac12(10)(4)=20\ \mathrm{m}, so it is at height 80−20=60 m80-20 = 60\ \mathrm{m}. Check with the second stone: 40(2)−12(10)(4)=60 m40(2)-\tfrac12(10)(4)=60\ \mathrm{m}. The shortcut is that the gravity terms cancel in the relative motion.
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Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake