The distance travelled by a body during last second of its total flight is d when the body is projected vertically up with certain velocity. If the velocity of projection is doubled, the distance travelled by the body during last second of its total flight is

Options

A. 2d
B. d
C. `2d+g/2`
D. `2d-g/2`

Answer

To solve the problem, we need to determine the distance traveled by a body during the last second of its flight when the initial velocity is doubled. Let's break down the solution step by step.

### Step 1: Understand the problem
The distance traveled during the last second of the flight of a body projected vertically can be calculated using the formula for the distance traveled in the nth second. The formula is:

\[ d_n = u + \frac{a}{2} (2n - 1) \]

where:
- \(d_n\) is the distance traveled during the nth second,
- \(u\) is the initial velocity,
- \(a\) is the acceleration (which is \(-g\) for upward motion),
- \(n\) is the total time of flight in seconds.

### Step 2: Calculate the time of flight for the initial velocity
For the initial velocity \(u\), the time of flight \(T\) can be calculated using:

\[ T = \frac{2u}{g} \]

### Step 3: Calculate the distance during the last second for the initial velocity
The distance traveled during the last second of flight when the initial velocity is \(u\) is:

\[ d = u + \frac{-g}{2} \left(2 \cdot \frac{2u}{g} - 1\right) \]

This simplifies to:

\[ d = u - \frac{g}{2} \left(\frac{4u}{g} - 1\right) = u - 2u + \frac{g}{2} = -u + \frac{g}{2} \]

### Step 4: Calculate the time of flight for the doubled initial velocity
Now, if the initial velocity is doubled to \(2u\), the new time of flight \(T'\) becomes:

\[ T' = \frac{2(2u)}{g} = \frac{4u}{g} \]

### Step 5: Calculate the distance during the last second for the doubled velocity
Now, we can find the distance traveled during the last second of flight for the doubled initial velocity:

\[ d' = 2u + \frac{-g}{2} \left(2 \cdot \frac{4u}{g} - 1\right) \]

This simplifies to:

\[ d' = 2u - \frac{g}{2} \left(\frac{8u}{g} - 1\right) = 2u - 4u + \frac{g}{2} = -2u + \frac{g}{2} \]

### Step 6: Relate the new distance to the original distance
From the original distance \(d = -u + \frac{g}{2}\), we can express \(d'\) in terms of \(d\):

\[ d' = 2(-u + \frac{g}{2}) = 2d \]

### Conclusion
Thus, when the initial velocity is doubled, the distance traveled by the body during the last second of its total flight is:

\[ d' = 2d \]

### Final Answer
The distance traveled by the body during the last second of its total flight when the velocity of projection is doubled is \(2d\).

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