A body thrown vertically upwards with an initial valocity `u` reaches maximum height in 6 s. The ratio of the distances traveled by the body in the first second the seventh second is

Options

A. `1:1`
B. `11:1`
C. `1:2`
D. `1:11`

Explanation

To solve the problem, we need to find the distances traveled by the body in the first second and the seventh second after it is thrown vertically upwards, and then calculate the ratio of these distances.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The body is thrown upwards with an initial velocity \( u \) and reaches its maximum height in 6 seconds. The time taken to reach maximum height is given as \( t = 6 \, \text{s} \).

2. **Finding the Initial Velocity**:
At maximum height, the final velocity \( v = 0 \). We can use the equation of motion:
\[ v = u - gt \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)). Plugging in the values:
\[ 0 = u - 10 \times 6 \]
\[ u = 60 \, \text{m/s} \]

3. **Distance Traveled in the First Second**:
To find the distance traveled in the first second, we use the equation:
\[ s_1 = ut + \frac{1}{2}gt^2 \]
For \( t = 1 \, \text{s} \):
\[ s_1 = 60 \times 1 + \frac{1}{2} \times (-10) \times (1)^2 \]
\[ s_1 = 60 - 5 = 55 \, \text{m} \]

4. **Distance Traveled in the Seventh Second**:
The seventh second refers to the time interval from \( t = 6 \, \text{s} \) to \( t = 7 \, \text{s} \). The distance traveled during this interval can be calculated using:
\[ s_7 = s(7) - s(6) \]
First, we find \( s(6) \) (the distance at \( t = 6 \)):
\[ s(6) = ut + \frac{1}{2}gt^2 = 60 \times 6 + \frac{1}{2} \times (-10) \times (6)^2 \]
\[ s(6) = 360 - 180 = 180 \, \text{m} \]
Now, we find \( s(7) \):
\[ s(7) = u \times 7 + \frac{1}{2} \times (-10) \times (7)^2 \]
\[ s(7) = 60 \times 7 + \frac{1}{2} \times (-10) \times (49) \]
\[ s(7) = 420 - 245 = 175 \, \text{m} \]
Now, we calculate \( s_7 \):
\[ s_7 = s(7) - s(6) = 175 - 180 = -5 \, \text{m} \]
Since we are looking for the distance traveled in the seventh second, we take the absolute value:
\[ s_7 = 5 \, \text{m} \]

5. **Calculating the Ratio**:
Now, we can find the ratio of the distances traveled in the first second and the seventh second:
\[ \text{Ratio} = \frac{s_1}{s_7} = \frac{55}{5} = 11 \]

### Final Answer:
The ratio of the distances traveled by the body in the first second to the seventh second is \( 11:1 \).