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A body is released from the top of a tower of height H metres. After 2 seconds it is stopped and then instantaneously released. What will be its height after next 2 seconds

after 2 second of releasing from the top
h=1/2gtsquare. u=0
=1/2*10*2*2. g=10m/s
=20m
then it is stopped and instantaneously released
there again velocity(v) bcomz 0
no similarly we get h=20m
now adding both we get 40m
dats the height of tower
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H-40
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h-20
 
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Do like this

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•In first 2 seconds, distance covered by the body is 20m ( using displacement formula). Hence remaining height to be covered is (H - 20) m. •Now, since it is stopped and instantaneously released, distance covered in next 2 seconds is 20 m (using displacement formula). Hence, the height after next 2 second is (H - 40)m.
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