A small object, of mass 0.02kg at the top of a building 160 m high, is dropped from rest.

Assuming that the air resistance has magnitude 0·096 N,

calculate the height of the object above the ground 4 s after it was dropped

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Here when the ball falls to the ground two forces will act on the ball. One is the weight of the ball which act downward which drives the ball to ground. In the upward direction air resistance will act on it.

Using Newtons second law of motion;

`darrF = ma`

`0.02xx9.81-0.096 = 0.02xxa`

`a = 5.01m/s^2`

Using the equations of motion;

`darrS = Ut+1/2at^2`

`S = 0+1/2xx5.01xx4^2`

`S = 40.08`

Height above ground `= 160-40.02 = 119.92`

*So the ball is 199.92m above ground when the time is 4s.*

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