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`        A force of magnitude 2mg(1-ay) starts acting in the vertically upward direction on a body of mass m placed on earth’s surface, where y is the height of the object above the ground during the ascent and ‘a’ is a positive constant. The total height through which the body ascends is a) 2/ab) 1/2ac) 2/3 ad) 1/a`
2 years ago

Arun
23345 Points
```							Dear Devansh In this case at max height eqillibrium will be there and in that condition,, Total downward force = Total upward force mg = 2mg ( 1 - ay ) 1 = 2 ( 1 - ay ) 1 = 2 - 2ay 2ay = 2 - 1 ay = 1/2 y = 1/2a   RegardsArun (askIITians forum expert)
```
2 years ago
ABHINAV Shrivastav
13 Points
```							Initial and final K.E of the body is zero Applying work energy principalIntegration of Fnet.dy from  0 to h must equals to zeroWhere fnet =F-mgBy putting value of fnet  We get [y-ay^2]from 0 to h =0Finally we get h=1/a.hence option D is correct
```
7 months ago
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