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Find the ratio of the distance moved by ...

Find the ratio of the distance moved by a free-falling body from rest in fourth and fifth seconds of its journey.

A

`4 :5`

B

`7:9`

C

`16:25`

D

`1:1`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the ratio of the distance moved by a free-falling body from rest in the fourth and fifth seconds of its journey, we can follow these steps: ### Step 1: Understand the motion of the free-falling body A free-falling body is one that is falling under the influence of gravity alone. The initial speed (U) is 0, and the acceleration (A) is equal to the acceleration due to gravity (g). ### Step 2: Use the formula for displacement in the nth second The displacement of a freely falling body during the nth second can be calculated using the formula: \[ S_n = U + \frac{A}{2} (2n - 1) \] Since the body is falling from rest, we have: - \( U = 0 \) - \( A = g \) Thus, the formula simplifies to: \[ S_n = \frac{g}{2} (2n - 1) \] ### Step 3: Calculate the distance moved in the 4th second (S4) For the 4th second (n = 4): \[ S_4 = \frac{g}{2} (2 \cdot 4 - 1) \] \[ S_4 = \frac{g}{2} (8 - 1) \] \[ S_4 = \frac{g}{2} \cdot 7 \] \[ S_4 = \frac{7g}{2} \] ### Step 4: Calculate the distance moved in the 5th second (S5) For the 5th second (n = 5): \[ S_5 = \frac{g}{2} (2 \cdot 5 - 1) \] \[ S_5 = \frac{g}{2} (10 - 1) \] \[ S_5 = \frac{g}{2} \cdot 9 \] \[ S_5 = \frac{9g}{2} \] ### Step 5: Find the ratio of S4 to S5 Now, we need to find the ratio \( \frac{S_4}{S_5} \): \[ \frac{S_4}{S_5} = \frac{\frac{7g}{2}}{\frac{9g}{2}} \] The \( \frac{g}{2} \) cancels out: \[ \frac{S_4}{S_5} = \frac{7}{9} \] ### Step 6: Write the final answer Thus, the ratio of the distance moved by the free-falling body in the fourth and fifth seconds is: \[ \text{Ratio} = 7:9 \]
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