A body is thrown vertically upwards with initial velocity 'u' reaches maximum height in 6 seconds. The ratio of distances travelled by the body in the first second and seventh second is

Options

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

Answer

To solve the problem step by step, we need to find the ratio of distances traveled by a body thrown vertically upwards in the first second and the seventh second. Here's how we can approach it:

### Step 1: Determine the initial velocity (u)
The body reaches its maximum height in 6 seconds. At maximum height, the final velocity (v) is 0. We can use the equation of motion:
\[ v = u + at \]
Where:
- \( v = 0 \) (at maximum height)
- \( a = -g = -10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 6 \, \text{s} \)

Substituting the values into the equation:
\[ 0 = u - 10 \times 6 \]
\[ u = 60 \, \text{m/s} \]

### Step 2: Calculate the distance traveled in the first second (s1)
The formula for the distance traveled in the nth second is:
\[ s_n = u + \frac{a}{2} (2n - 1) \]
For the first second (n = 1):
\[ s_1 = u + \frac{a}{2} (2 \times 1 - 1) \]
Substituting the values:
\[ s_1 = 60 + \frac{-10}{2} (1) \]
\[ s_1 = 60 - 5 = 55 \, \text{m} \]

### Step 3: Calculate the distance traveled in the seventh second (s7)
Using the same formula for n = 7:
\[ s_7 = u + \frac{a}{2} (2 \times 7 - 1) \]
Substituting the values:
\[ s_7 = 60 + \frac{-10}{2} (13) \]
\[ s_7 = 60 - 65 = -5 \, \text{m} \]

### Step 4: Calculate the ratio of distances
Now we can find the ratio of distances traveled in the first second to the seventh second:
\[ \text{Ratio} = \frac{s_1}{s_7} = \frac{55}{-5} = -11 \]

Since we are interested in the magnitude of the ratio:
\[ \text{Ratio} = 11:1 \]

### Final Answer
The ratio of distances traveled by the body in the first second and the seventh second is:
\[ \text{Ratio} = 11:1 \]

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