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 of air resistance]:-

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-by-Step Solution:

1. **Identify the known values:**
- Initial velocity (u) = 40 m/s
- Final velocity (v) at the maximum height = 0 m/s
- Acceleration due to gravity (g) = 9.8 m/s² (acting downwards)

2. **Calculate the maximum height (h) reached by the body:**
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Here, \(a = -g\) (since gravity acts downward), so:
\[ 0 = (40)^2 + 2(-9.8)(h) \]
Rearranging gives:
\[ 0 = 1600 - 19.6h \implies 19.6h = 1600 \implies h = \frac{1600}{19.6} \approx 81.63 \text{ meters} \]

3. **Calculate the total time (t) taken to reach the maximum height:**
We can use the equation:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting the known values:
\[ 81.63 = 40t - \frac{1}{2}(9.8)t^2 \]
This simplifies to:
\[ 81.63 = 40t - 4.9t^2 \]
Rearranging gives us a quadratic equation:
\[ 4.9t^2 - 40t + 81.63 = 0 \]

4. **Solve the quadratic equation for time (t):**
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 4.9\), \(b = -40\), and \(c = 81.63\).
\[ t = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 4.9 \cdot 81.63}}{2 \cdot 4.9} \]
Calculating the discriminant:
\[ = 1600 - 1601.56 \approx 0.56 \]
Thus,
\[ t = \frac{40 \pm \sqrt{0.56}}{9.8} \]
Approximating gives:
\[ t \approx 4.08 \text{ seconds} \]

5. **Calculate the distance traveled in the last second:**
The distance traveled in the last second can be calculated as:
\[ \text{Distance in last second} = \text{Distance at } t - \text{Distance at } (t-1) \]
Using the formula for distance:
\[ s = ut + \frac{1}{2}at^2 \]
For \(t = 4.08\):
\[ s(4.08) = 40(4.08) - \frac{1}{2}(9.8)(4.08)^2 \]
For \(t = 3.08\):
\[ s(3.08) = 40(3.08) - \frac{1}{2}(9.8)(3.08)^2 \]
Subtracting gives the distance in the last second.

6. **Final Calculation:**
After performing the calculations, we find:
\[ \text{Distance in last second} \approx 4.9 \text{ meters} \]

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