A ball is released from the top of a tower. The ratio of work done by force of gravity in 1st second, 2nd second and 3rd second of the motion of ball is

Options

A. `1 : 2 : 3`
B. `1 : 4 : 16`
C. `1 : 3 : 5`
D. `1 : 9 : 25`

Answer

To solve the problem of finding the ratio of work done by the force of gravity in the 1st, 2nd, and 3rd seconds of the motion of a ball released from the top of a tower, we can follow these steps:

### Step 1: Understand the Concept of Work Done
Work done (W) by a force is defined as the product of the force (F) and the displacement (s) in the direction of the force. Mathematically, it can be expressed as:
\[ W = F \cdot s = F \cdot s \cdot \cos(\theta) \]
where \( \theta \) is the angle between the force and displacement vectors. In this case, since the force of gravity acts downward and the displacement is also downward, \( \theta = 0^\circ \) and \( \cos(0) = 1 \).

### Step 2: Calculate Displacement in Each Second
The displacement of the ball during the nth second can be calculated using the formula:
\[ s_n = u + \frac{a}{2} \cdot (2n - 1) \]
where:
- \( u \) is the initial velocity (which is 0 since the ball is released),
- \( a \) is the acceleration (which is \( g \), the acceleration due to gravity),
- \( n \) is the second for which we are calculating the displacement.

#### For the 1st Second (n=1):
\[ s_1 = 0 + \frac{g}{2} \cdot (2 \cdot 1 - 1) = \frac{g}{2} \cdot 1 = \frac{g}{2} \]

#### For the 2nd Second (n=2):
\[ s_2 = 0 + \frac{g}{2} \cdot (2 \cdot 2 - 1) = \frac{g}{2} \cdot 3 = \frac{3g}{2} \]

#### For the 3rd Second (n=3):
\[ s_3 = 0 + \frac{g}{2} \cdot (2 \cdot 3 - 1) = \frac{g}{2} \cdot 5 = \frac{5g}{2} \]

### Step 3: Calculate Work Done in Each Second
Now we can calculate the work done by the force of gravity for each second.

#### Work Done in the 1st Second:
\[ W_1 = mg \cdot s_1 = mg \cdot \frac{g}{2} = \frac{mg^2}{2} \]

#### Work Done in the 2nd Second:
\[ W_2 = mg \cdot s_2 = mg \cdot \frac{3g}{2} = \frac{3mg^2}{2} \]

#### Work Done in the 3rd Second:
\[ W_3 = mg \cdot s_3 = mg \cdot \frac{5g}{2} = \frac{5mg^2}{2} \]

### Step 4: Find the Ratio of Work Done
Now we can find the ratio of work done in the 1st, 2nd, and 3rd seconds:
\[ \text{Ratio} = W_1 : W_2 : W_3 = \frac{mg^2}{2} : \frac{3mg^2}{2} : \frac{5mg^2}{2} \]

This simplifies to:
\[ 1 : 3 : 5 \]

### Final Answer
The ratio of work done by the force of gravity in the 1st second, 2nd second, and 3rd second of the motion of the ball is:
\[ \boxed{1 : 3 : 5} \]