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 motion of the ball
When the ball is released from rest, it falls freely under the influence of gravity. The distance it travels in each second can be calculated using the equations of motion.

### Step 2: Calculate the displacement in each second
The displacement \( S_n \) during the \( n \)-th second can be calculated using the formula:
\[ S_n = g \cdot n - \frac{1}{2} g \]
where \( g \) is the acceleration due to gravity.

#### For the 1st second (\( n = 1 \)):
\[ S_1 = g \cdot 1 - \frac{1}{2} g = g - \frac{1}{2} g = \frac{g}{2} \]

#### For the 2nd second (\( n = 2 \)):
\[ S_2 = g \cdot 2 - \frac{1}{2} g = 2g - \frac{1}{2} g = \frac{3g}{2} \]

#### For the 3rd second (\( n = 3 \)):
\[ S_3 = g \cdot 3 - \frac{1}{2} g = 3g - \frac{1}{2} g = \frac{5g}{2} \]

### Step 3: Write the displacements
Now we have the displacements for each second:
- \( S_1 = \frac{g}{2} \)
- \( S_2 = \frac{3g}{2} \)
- \( S_3 = \frac{5g}{2} \)

### Step 4: Find the ratio of displacements
The ratio of displacements \( S_1 : S_2 : S_3 \) is:
\[ S_1 : S_2 : S_3 = \frac{g}{2} : \frac{3g}{2} : \frac{5g}{2} \]
Cancelling \( \frac{g}{2} \) from each term gives:
\[ 1 : 3 : 5 \]

### Step 5: Calculate the work done by gravity
The work done by gravity in each second can be calculated using the formula:
\[ \text{Work} = \text{Force} \times \text{Displacement} \]
Since the force of gravity (weight of the ball) remains constant, the work done in each second will be directly proportional to the displacement in that second.

### Step 6: Write the ratio of work done
Thus, the ratio of work done by gravity in the 1st, 2nd, and 3rd seconds will also be:
\[ \text{Work}_1 : \text{Work}_2 : \text{Work}_3 = 1 : 3 : 5 \]

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