### Free fall, Acceleration due to Gravity and Kinetic Energy

##### March 31, 2017

Weight of any object is a force applied on the object by earth’s gravitational pull towards the center of the earth. We know the equation of a** force **= mass * acceleration.

###### i.e. F=m*a…….(1)

When we talk of weight force then the equation becomes like this:

###### Weight(W)= mass * acceleration due to gravity = m * g……(2)

The value of g on earth is nearly 9.8 meter/second square.

This g varies a bit on different places of earth, but that’s a different story. [read here:**Acceleration due to gravity and its variation due to height and depth** ]

For now we will stick to the value given above or for simplicity of understanding and quick calculation we may take this as 10 m/sec (disclaimer:For exam follow instructions given in your Question paper for this value)

**zero air resistance during free fall**

As an object experiences free fall **(considering zero air resistance)**, the downward force acting on it is its weight itself (mg). As we are considering zero air resistance, there is no drag force and as a result no Terminal velocity will be attained. The object will fall under the influence of a single force, i.e. Gravity.

This is caused by the earth’s pull on the object towards the center of the earth.

Under this force it’s velocity constantly rises till it touches the ground. That means the objects experiences acceleration which is known in this case as ‘Acceleration due to gravity’.

From a formula of linear motion, Current Velocity(V) = Initial Velocity (U) + Acceleration due to gravity(g)*Time elapsed (t)

###### V=U + gt………………………….(3)

Say a stone is being dropped from the roof of tall building. As it’s static before being dropped, so we can take the initial velocity(U) as zero.

Let’s take the value of g as 10 in SI for simplicity of calculation.

So after 1 sec of dropping, when the stone is freely falling, it’s velocity towards downwards would be

V = 0 + 10*1 m/sec

or V=10 m/s

Again after 2 seconds of dropping, the velocity becomes ,

V= 0 + 10*2 = 20 m/sec. (Pls note that the initial velocity is 0 similar to the first case above because this velocity is the velocity of the stone before the freefall starts and as it was static at that time so we have to consider it’s value as 0 for such cases)

So it’s clear that after every second of free fall, the stone is gaining an additional velocity of 10 m/s. So after 3 seconds of dropping its velocity would be 30 m/sec and after 4 seconds of dropping it will be 40 m/sec.

So it’s evident that if this stone stone remains in air for 10 seconds before it touches the ground then just before touching the ground it’s velocity would be 100 meter/second.**(considering zero air resistance)**

Kinetic energy is the energy possessed by a body because of its velocity.

###### feel weightless during freefall?

Here one question arises. Why does a person having a free fall (no air resistance or any drag force against his downward motion) feel weightless?

This is because the feeling of weight comes when a normal reaction force is exerted on our body by the surface on which we stand or rest.

While freefall (without any air resistance), there is no such surface that can provide this normal reaction, as a result the feeling of weight vanishes.

The person (with mass m) on freefall **feels** weightless as it doesn’t experience any normal reaction.. But remember that a force equal to mg is still working on him to pull him towards the centre of the earth. For details see this post. weightlessness

###### Formula of Kinetic Energy (KE)= ½ mv^{2 }

where m is the mass and v is the velocity of the body.

When a body is freely falling, at any point in time during its transit, before touching ground it will have Kinetic Energy.

As well as it will have a Potential Energy while it’s in air due to its height from the ground.

We can show that the sum of potential energy and kinetic energy during this transit at any point will remain constant.

When air resistance is considerable and come into play, we have to consider the air drag force which acts opposite to the weight of the falling body. See a post Free Fall against air drag

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