When an object is dropped from some height, its velocity increases at a constant rate. In other words, when an object is dropped from some height, a uniform acceleration is produced in it by the gravitational pull of the earth and this acceleration due to gravity is denoted by letter ' g' .Thus, acceleration due to gravity , g = 9.8 `(m)/(s^2)`.

When a body is dropped freely, it falls with an acceleration of 9.8 `(m)/(s^2)` and when a body is thrown vertically upwards, it undergoes retardation of 9.8 `(m)/(s^2)` . This is the value of acceleration due to gravity.

- Acceleration due to gravity (g) does not depend on the Mass of a body.
- The laws of motion for freely falling bodies - the acceleration 'a' is changed to 'g' and distance 's' is changed to height 'h'

(i) v = u +at changes to v = u + gt

(ii) s = ut + `(1)/(2)` at^{2} changes to h = ut + `(1)/(2)` gt^{2}

(iii) v ^{2} = u^{2 }+ 2as changes to v^{2} = u^{2} + 2gh

These are the equations for acceleration due to gravity. The acceleration can be found with the help of the acceleration due to gravity formulas.

We should remember the following points to solve numerical problems based on freely falling objects:

- The acceleration due of gravity of earth (g) is always taken as negative and written with a minus sign (irrespective of whether a body is thrown upwards or it falls downwards)
- When a body is dropped freely from a height , its initial velocity 'u' is zero
- When a body is thrown vertically upwards, its final velocity 'v' becomes zero
- The time taken by a body to rise to the height point is equal top the time taken to fall from the same height.

My Upcoming post is on **Center of Mass vs Center of Gravity** keep checking my blogs.

Listed below are some of the examples on acceleration due to gravity.

**Example 1:**

To estimate the height of the bridge over a river, a stone is dropped freely in the river from the bridge. The stone takes 2 seconds to touch the water surface in the river. Calculate the height of the bridge from the water level (g = 9.8 `(m)/(s^2)` )

**Solution:**

The stone is being dropped freely from rest, so the initial velocity of the stone is u = 0.

The acceleration due to gravity g is taken as negative and written with a minus sign

Initial velocity of stone, u = 0

Acceleration due to gravity, g = - 9.8 `(m)/(s^2)` .

Time taken, t = 2 sec

To find the height of the bridge h="?

The height is given by the formula h = ut + `(1)/(2)` gt^{2}

Plugging in the values, we calculate the height

h^{ }= 0 * 2 + `(1)/(2)` (-9.8 )(2)^{2}

h = - `(1)/(2)` * 9.8 * 4 = - 19.6 m

Thus, the height of the bridge above the water level is 19.6 metres. The minus sign with the height shows that it is in the downward direction.

**Example 2:**

A ball is thrown up with a speed of 15 `(m)/(s)` . How high will it go before it begins to fall ? ( g="9.8" `(m)/(s^2)` ).

**Solution:**

Initial speed of ball, u = 15 `(m)/(s)`

Final speed of ball , v = 0 ( The ball stops)

Acceleration due to gravity g = - 9.8 `(m)/(s^2)` .

We need to find the height.

Formula used is v^{2} = u^{2}+ 2gh

Plugging in all the values, we get

v^{2} = u^{2}+ 2gh

(0)^{2} = (15)^{2} + 2(-9.8)h

0 = 225 -19.6 h

19.6 h = 225

h = `(225)/(19.6)` = 11.4 m

Having problem with **definition of angular momentum** keep reading my upcoming posts, i will try to help you.

**Example 3:**

A cricket ball is dropped from a height of 20 metres.

(a) Calculate the speed of the ball when it hits the ground

(b) Calculate the time it takes to fall through this height (g = 10 `(m)/(s^2)` .)

**Solution:**

Initial speed u = 0

Acceleration due to gravity = 10`(m)/(s^2)` .

Height, h = 20 m

We need to find the final speed v first

v^{2} = u^{2}+ 2gh

Plugging in all the values, we get

v^{2} = (0)^{2} + 2( -10) * 20

v^{2} = - 400 taking square roots on both sides

v = - `sqrt(400)`

v = -20 `(m)/(s)`

Thus, the speed of the cricket ball when it hits the ground is 20 metres per second.The minus sign with the speed shows that it is in the downward direction.

Now, we need to calculate the time taken t

Formula used is v = u + gt

Plugging in the values, we get

-20 = 0 + (-10) t

-20 = -10 t

10 t = 20

t = `(20)/(10)` = 2 s

Thus, the ball takes 2 second to fall from a height of 20 metres.