An aeroplane flying horizontally with a speed of 360 km `h^(-1)` releases a bomb at a height of 490 m from the ground. If g = 9. 8 m `s^(-2)` , it will strike the ground at

Options

A. 10 km
B. 100 km
C. 1 km
D. 16 km

Explanation

To solve the problem of the bomb released from an airplane, we will follow these steps:

### Step 1: Convert the speed of the airplane from km/h to m/s
The speed of the airplane is given as 360 km/h. To convert this to meters per second (m/s), we use the conversion factor:
\[ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \]
So,
\[ \text{Speed in m/s} = 360 \text{ km/h} \times \frac{1}{3.6} = 100 \text{ m/s} \]

### Step 2: Calculate the time taken for the bomb to fall 490 m
We will use the second equation of motion to find the time taken for the bomb to fall. The equation is:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \(s\) = vertical distance fallen (490 m)
- \(u\) = initial vertical velocity (0 m/s, since the bomb is released)
- \(a\) = acceleration due to gravity (9.8 m/s²)
- \(t\) = time in seconds

Substituting the values into the equation:
\[ 490 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
This simplifies to:
\[ 490 = 4.9 t^2 \]
Now, solving for \(t^2\):
\[ t^2 = \frac{490}{4.9} = 100 \]
Taking the square root:
\[ t = \sqrt{100} = 10 \text{ seconds} \]

### Step 3: Calculate the horizontal distance traveled by the bomb
The horizontal distance \(d\) can be calculated using the formula:
\[ d = vt \]
Where:
- \(v\) = horizontal speed of the airplane (100 m/s)
- \(t\) = time taken to fall (10 s)

Substituting the values:
\[ d = 100 \text{ m/s} \times 10 \text{ s} = 1000 \text{ m} \]

### Step 4: Convert the distance from meters to kilometers
Since we need the answer in kilometers:
\[ d = \frac{1000 \text{ m}}{1000} = 1 \text{ km} \]

### Final Answer
The bomb will strike the ground at a distance of **1 km** from the point directly below where it was released.

---