a man in a balloon rising vertically with an accleration of 4.9m/s releases a ball 2 seconds after the balloon is let go from the ground . the greatest height above the ground reached by the ball is??

Asked by MRUGAKSHI DOSHI | 2nd Nov, 2010, 07:12: PM

Expert Answer:

Dear Student,

a = 4.9 m/s2, initial velocity = 0

Velocity of balloon and ball after 2 seconds

v = u + at

   = 4.9 * 2

   = 9.8 m/s

 

Height at this point

h = 1/2 at2

   = 0.5 * 4.9 * 4

   = 9.8 m

 

The ball will attain further height = s given by

1/2 mv2 = mgs

s = v2/2g

  =  4.9 m

 

Hence greatest height reached by the ball above the ground

H = h + s

    = 9.8 + 4.9

    = 14.7 m  

We hope that clarifies your query.
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Answered by  | 3rd Nov, 2010, 12:17: AM

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