CBSE Class 11-science Answered
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 | 02 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.
Regards
Team
Topperlearning
Answered by | 03 Nov, 2010, 12:17: AM
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