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A ball is allowed to fall from the top of tower of 200m high. At the same instant another ball is thrown vertically upwards from the bottom of the tower

with a velocity of 40m/s. When and where the two balls meet?



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The position of the ball from the top of the tower after time  t  is
   h1 = ut + 1/2 g t²  = 0 + 1/2 * g * t²

  height above ground of the ball thrown from ground - after time  t  is

   h2  =  u t - 1/2 g t²  ,          u = 40 m/sec

   h1 + h2 = 200 meters

     u t - 1/2 g t² + 1/2 g t²  =  200 meters
     t = 200 / u = 5 seconds

     h2 = height above ground = 200 - 1/2 g t² = 75 meters

2 5 2
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