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A 25-newton weight falls freely from rest from

the roof of a building. What is the total distance

the weight falls in the first 1.0 second?

(1) 19.6 m (3) 4.9 m

(2) 9.8 m (4) 2.5 m

Verified Answer :

AL2006


The acceleration of gravity on Earth is 9.8 m/s² .

Neglecting air resistance, an object starting at rest
is falling 9.8 m/s after 1 second.

Its average speed during that second is  (1/2) (0 + 9.8 m/s)

                                                                 =  4.9 m/s .

So it falls 4.9 meters during that second.

The weight of the object doesn't enter in to this calculation.
The acceleration of gravity is the acceleration of gravity, and
the weight of the object doesn't matter.  If you can find a
chamber that's big enough and you pump all the air out of it,
then you can drop a feather and a battleship in the chamber,
and they fall together and hit the floor at the same time. 

Answer:

Distance covered, d = 4.9 meters

Explanation:

It is given that,

Force acting on the freely falling object, F = 25 N

Let d is the distance  the weight falls in the first 1.0 second. It can be calculated using the second equation of motion as :

\(d=ut+\dfrac{1}{2}gt^2\)

u = 0 here

\(d=\dfrac{1}{2}\times 9.8\ m/s^2\times (1\ s)^2\)

d = 4.9 meters

So, the distance  the weight falls in the first 1.0 second is 4.9 meters. Hence, this is the required solution.

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