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JEE Main 2014 · Motion Under Gravity

Motion Under Gravity question from JEE Main 2014

Q47JEE Main 2014Medium

Motion Under Gravity

From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:

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Answer: (c)

  1. Time to reach the highest point: t0=ugt_0 = \frac{u}{g}, so time to hit the ground t=nugt = \frac{nu}{g}.
  2. Upward positive, origin at top of tower: −H=ut−12gt2-H = ut - \frac{1}{2}gt^2
  3. −H=nu2g−n2u22g⇒2gH=n2u2−2nu2-H = \frac{nu^2}{g} - \frac{n^2u^2}{2g} \Rightarrow 2gH = n^2u^2 - 2nu^2
  4. So the answer is (c) 2gH=nu2(n−2)2gH = nu^2(n-2).
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