A body is projected vertically upward with speed 40m/s. The distance travelled by body in the last second of upward journey is [take `g=9.8m/s^(2)` and neglect eggect of air rresistance]:-

Options

A. 4.9 m
B. 9.8 m
C. 12.4 m
D. 19.6 m

Answer

To solve the problem of finding the distance traveled by a body in the last second of its upward journey when projected vertically upward with an initial speed of 40 m/s, we can follow these steps:

### Step 1: Calculate the maximum height (h) reached by the body.
We can use the formula for maximum height reached when a body is projected upwards:
\[ h = \frac{u^2}{2g} \]
Where:
- \( u = 40 \, \text{m/s} \) (initial velocity)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ h = \frac{(40)^2}{2 \times 9.8} = \frac{1600}{19.6} \approx 81.63 \, \text{m} \]

### Step 2: Calculate the total time (T) taken to reach the maximum height.
Using the formula:
\[ T = \frac{u}{g} \]
Substituting the values:
\[ T = \frac{40}{9.8} \approx 4.08 \, \text{s} \]

### Step 3: Calculate the distance traveled in the last second.
To find the distance traveled in the last second, we need to find the distance traveled in the time interval from \( T-1 \) to \( T \).

Using the formula for distance traveled:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( u = 40 \, \text{m/s} \)
- \( a = -g = -9.8 \, \text{m/s}^2 \) (deceleration due to gravity)
- \( t = T - 1 = 4.08 - 1 = 3.08 \, \text{s} \)

Now, substituting the values into the equation:
\[ s = 40 \times 3.08 + \frac{1}{2} \times (-9.8) \times (3.08)^2 \]

Calculating each term:
1. \( 40 \times 3.08 = 123.2 \, \text{m} \)
2. \( \frac{1}{2} \times (-9.8) \times (3.08)^2 = -4.9 \times 9.4864 \approx -46.4 \, \text{m} \)

Now, substituting these values back:
\[ s = 123.2 - 46.4 = 76.8 \, \text{m} \]

### Step 4: Calculate the distance traveled in the last second.
The distance traveled in the last second is:
\[ \text{Distance in last second} = h - s \]
Where \( h = 81.63 \, \text{m} \) and \( s = 76.8 \, \text{m} \):
\[ \text{Distance in last second} = 81.63 - 76.8 = 4.83 \, \text{m} \]

### Final Answer:
The distance traveled by the body in the last second of its upward journey is approximately **4.9 m**.

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