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Assume that an object of mass 'm' is falling to the earth.

The force acting on the same would be F = m*g = 10m (assuming g = 10m/s^2).
In this case the velocity of the object at time 't' = 10 seconds will be 100 m/s (v = u + at).
At this point, if an opposing force of F' = -10m is applied on the falling object, how long would it take for the velocity of the falling object to reduce to 0?

Would it be valid to say that the opposing force of F' = -10m has to be applied for time 't' = 10 seconds to reduce the velocity of the falling object to 0?

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1 Answer 1

up vote 2 down vote accepted

At the moment the force F' starts to act, you will still have gravity. There is no net force, thus no deceleration.

If you this opposing force is indeed the only force, it will indeed take 10 seconds to reach zero velocity, since now you get $v=u_0 -a t$ And $u_0=a T$

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If the opposing force F' = 20m then the net force acting upwards would be F' - F = -10m. In this case there would be deceleration and in 10 seconds the velocity will reach 0. Kindly advise if this is valid. –  Ram Sidharth Feb 10 '12 at 17:57
    
@RamSidharth That is indeed correct –  Bernhard Feb 10 '12 at 22:27
    
Thanks a lot for the response, Bernhard. –  Ram Sidharth Feb 11 '12 at 18:26

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