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AP EAMCET201823 Apr 2018Morning ShiftPhysicsMotion in One DimensionActual

A body is projected vertically upwards with a velocity u from the top of a tower. Time taker by it to reach the ground is n times, then the time taken by it to reach the highest point in its path. Height of the tower is

Options

  1. An u^2(n-1) 2 g
  2. Bn u^2(n-2) g
  3. Cn u^2(n-2) 2 g
  4. Du^2 2 g (n+1)

Correct answer

C. n u^2(n-2) 2 g

Step-by-step solution

Let the time taken to reach the maximum height, when thrown vertically upwards t₁= u g If t₂ be the time to hit the ground, then given t₂=n t₁=n u / g aligned & = | 2 x 1+ 2 x | Now, IF & =e^ | 2 x 1+ 2 x | & = 2 x 1+ 2 x = 2 x 2 ^2 x aligned Now, solution of differential equation is aligned & y. IF = ( . IF ) d x & y 2 x 2 ^2 x = ^2 x 2 x 2 ^2 x d x & y ( 2 x ^2 x )= 2 x d x & y ( 1- ^2 x 1+ ^2 x ) 1 ^2 x = 2 x 2 +c₁ & y (1- ^2 x ) ^2 x ^2 x = 2 x 2 +c₁ & y (1- ^2 x )= 2 x+2 c₁ 2 & y= 2 x+c 2 (1- ^2 x ) & aligned

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