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JEE Main 2017 · Physics · Kinematics and Laws of Motion · medium

A stone is projected vertically upward with speed 20 m s^-1 from the top of a tower 25 m high. Taking g = 10 m s^-2 and neglecting air resistance…

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A stone is projected vertically upward with speed 20 m s−120\ \mathrm{m\,s^{-1}} from the top of a tower 25 m25\ \mathrm{m} high. Taking g=10 m s−2g = 10\ \mathrm{m\,s^{-2}} and neglecting air resistance, the average speed of the stone from projection until it strikes the ground is
  1. A. 5 m s−15\ \mathrm{m\,s^{-1}}
  2. B. 9 m s−19\ \mathrm{m\,s^{-1}}
  3. C. 13 m s−113\ \mathrm{m\,s^{-1}}
  4. D. 15 m s−115\ \mathrm{m\,s^{-1}}
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Answer: (C) 13 m s−113\ \mathrm{m\,s^{-1}}

Taking up as positive and the ground as −25 m-25\ \mathrm{m}: −25=20t−5t2-25 = 20t - 5t^2, so t2−4t−5=0t^2 - 4t - 5 = 0 and t=5 st = 5\ \mathrm{s}. The stone rises h=202/(2×10)=20 mh = 20^2/(2\times 10) = 20\ \mathrm{m} and then falls 20+25=45 m20 + 25 = 45\ \mathrm{m}, so the total path length is 65 m65\ \mathrm{m}. Average speed =65/5=13 m s−1= 65/5 = 13\ \mathrm{m\,s^{-1}}. The value 5 m s−15\ \mathrm{m\,s^{-1}} is the magnitude of the average velocity (25/525/5), which is a different quantity because displacement is not path length.
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