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## Equation of Motion for a Body Falling Under Gravity

Since the freely falling bodies fall with uniformly accelerated motion, the three equations of motion derived earlier for bodies under uniform acceleration can be applied to the motion of freely falling bodies. For freely falling bodies, the acceleration due to gravity is ‘g’, so we replace the acceleration V of the equations by ‘g’ and since the vertical distance of the freely falling bodies is known as height ‘h’, we replace the distance ‘s’ in our equations by the height ‘h’. This gives us the following modified equations for the motion of freely falling bodies :

We will use these modified equations to solve numerical problems. Before we do that, we should remember the following important points for the motion of freely falling bodies :

(a) When a body is falling vertically downwards, its velocity is increasing, so the acceleration due to gravity, g, is taken as positive. That is, Acceleration due to gravity = + 9.8 m/s^{2 }for a freely falling body

(b) When a body is thrown vertically upwards, its velocity is decreasing, so the acceleration due to gravity, g, is taken as negative. That is, Acceleration due to gravity = – 9.8 m/s^{2 }for a body thrown upwards

(c) When a body is dropped freely from a height, its initial velocity V is zero.

(d) When a body is thrown vertically upwards, its final velocity V becomes zero.

(e) The time taken by a body to rise to the highest point is equal to the time it takes to fall from the same height.

We will now solve some problems based on the motion of freely falling bodies.

**Example Problem 1.**

To estimate the height of a 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, u = 0. Again, the velocity of stone is increasing as it comes down, so the acceleration due to gravity, g, is to be taken as positive.

Now, Initial velocity of stone, u = 0

Time taken, t = 2 s

Acceleration due to gravity, g = 9.8 m/s^{2} (Stone comes down)

And, Height of the bridge, h = ? (To be calculated)

We know that for a freely falling body :

Height, h = ut + \(\frac{1}{2}\)gt^{2}

Putting the above values in this formula, we get:

h = 0 × 2 + \(\frac{1}{2}\) × 9.8 × (2)^{2}

or h = \(\frac{1}{2}\) × 9.8 × 4

or h = 19.6 m

Thus, the height of bridge above the water level is 19.6 metres.

**Example Problem 2.**

When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to gravity, g = 9.8 m/s^{2})

**Solution.**

Here, the ball is going up against the attraction of earth, so its velocity is decreasing continuously. In other words, we can say that the ball is being retarded. Thus, the acceleration in the ball is negative which means that the value of g is to be used here with the negative sign.

Here, Initial velocity of ball, u = ? (To be calculated)

Final velocity of ball, v = 0 (It stops)

Acceleration due to gravity, y = -9.8m/s^{2} (Ball goes up)

And, Height, h = 19.6 m

Now, putting all these values in the formula :

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

we get: (0)^{2} = u^{2} + 2 × (- 9.8) × 19.6

0 = u^{2} – 19.6 × 19.6

u^{2} = (19.6)^{2}

So, u = 19.6 m/s

Thus, the initial velocity of the ball is 19.6 m/s which means that the ball has been thrown upwards with a velocity of 19.6 m/s.

Let us now calculate the time taken by the ball to reach the highest point. Now, we know the initial velocity, the final velocity and the acceleration due to gravity, so the time taken can be calculated by using the equation:

v = u + gt

Here, Final velocity, v = 0 (The ball stops)

Initial velocity, u = 19.6 m/s (Calculated above)

Acceleration due to gravity, g = – 9.8 m/s^{2} (Ball goes up)

And, Time, t = ? (To be calculated)

So, putting these values in the above equation, we get:

0 = 19.6 + (- 9.8) × t

0 = 19.6 – 9.8 t

9.8 t = 19.6

t = \(\frac{19.6}{9.8}\)

t = 2 s

Thus, the ball takes 2 seconds to reach the highest point of its upward journey. Please note that the ball will take an equal time, that is, 2 seconds to fall back to the ground. In other words, the ball will take a total of 2 + 2 = 4 seconds to reach back to the thrower.

**Example Problem 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.**

(a) Here, Initial speed, u = 0

Final speed, v = ? (To be calculated)

Acceleration due to gravity, g = 10 m/s^{2} (Ball comes down)

And, Height, h = 20 m

Now, we know that for a freely falling body :

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

So, v^{2} = (0)^{2} + 2 × 10 × 20

v^{2} = 400

v = \(\sqrt{400}\)

or v = 20 m/s

Thus, the speed of cricket ball when it hits the ground will be 20 metres per second.

(b) Now, Initial speed, u = 0

Final speed, v = 20 m/s (Calculated above)

Acceleration due to gravity, g = 10 m/s^{2}

And, Time, t = ? (To be calculated)

Putting these Values in the formula :

v = u + gt,

we get: 20 = 0 + 10 × t

10t = 20

t = \(\frac{20}{10}\)

t = 2 s

Thus, the ball takes 2 seconds to fall through a height of 20 metres.

**Example Problem 4.**

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.**

Please note that here the ball is going up against the gravity, so the value of g is to be taken as negative.

Here, 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} (Retardation)

And, Height, h = ? (To be calculated)

Now, putting all these values in the formula,

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

we get: (0)^{2} = (15)^{2} + 2 × (- 9.8) × h

0 = 225 – 19.6 h

19.6 h = 225

or h = \(\frac{225}{19.6}\)

h = 11.4 m

Thus, the ball will go to a maximum height of 11.4 metres before it begins to fall.