The distance travelled by a falling body in the last second of its motion, to that in the last but one second is 7: 5, the velocity with which body strikes the ground is

Options

A. `19.6m//s`
B. `39.2 m//s`
C. `29.4 m//s`
D. `49 m//s`

Answer

To solve the problem, we need to find the velocity with which a falling body strikes the ground, given the ratio of distances traveled in the last second and the last but one second of its motion.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We are given that the distance traveled by a falling body in the last second of its motion (let's denote it as \( S_n \)) to that in the last but one second (denote it as \( S_{n-1} \)) is in the ratio 7:5.

2. **Using the Formula for Distance:**
The distance traveled in the nth second can be expressed as:
\[ S_n = u + \frac{1}{2} g (n^2) - \left( u + \frac{1}{2} g ((n-1)^2) \right) = \frac{1}{2} g (2n - 1) \]
Since the initial velocity \( u = 0 \) for a freely falling body, we simplify this to:
\[ S_n = \frac{1}{2} g (2n - 1) \]
Similarly, for \( S_{n-1} \):
\[ S_{n-1} = \frac{1}{2} g (2(n-1) - 1) = \frac{1}{2} g (2n - 3) \]

3. **Setting Up the Ratio:**
Given the ratio \( \frac{S_n}{S_{n-1}} = \frac{7}{5} \):
\[ \frac{\frac{1}{2} g (2n - 1)}{\frac{1}{2} g (2n - 3)} = \frac{7}{5} \]
The \( \frac{1}{2} g \) cancels out, leading to:
\[ \frac{2n - 1}{2n - 3} = \frac{7}{5} \]

4. **Cross-Multiplying:**
Cross-multiplying gives:
\[ 5(2n - 1) = 7(2n - 3) \]
Expanding both sides:
\[ 10n - 5 = 14n - 21 \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ 21 - 5 = 14n - 10n \]
Thus:
\[ 16 = 4n \implies n = 4 \]

6. **Finding the Final Velocity:**
Now that we have \( n = 4 \), we can find the final velocity \( v \) using the equation:
\[ v = u + gt \]
Since \( u = 0 \):
\[ v = gt = g \cdot 4 \]
Assuming \( g \approx 9.8 \, \text{m/s}^2 \):
\[ v = 9.8 \cdot 4 = 39.2 \, \text{m/s} \]

### Final Answer:
The velocity with which the body strikes the ground is \( 39.2 \, \text{m/s} \).