An acroplane is flying horizontally with a velocity of 600 km/h and a height of 1960m. When it is vectrically above a point A on the ground a bomb is released from it. The bomb strikes the ground at point B. the distance AB is:

Options

A. 1200m
B. 0.33km
C. 333.3km
D. 3.33km

Explanation

To solve the problem of finding the distance \( AB \) where a bomb released from an airplane strikes the ground, we can follow these steps:

### Step 1: Identify Given Data
- Horizontal velocity of the airplane, \( v = 600 \, \text{km/h} \)
- Height from which the bomb is released, \( h = 1960 \, \text{m} \)

### Step 2: Convert Horizontal Velocity to m/s
To work with consistent units, we convert the horizontal velocity from km/h to m/s:
\[ v = 600 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = \frac{600 \times 1000}{3600} \approx 166.67 \, \text{m/s} \]

### Step 3: Calculate Time of Flight
The time \( t \) it takes for the bomb to fall to the ground can be calculated using the equation of motion for vertical displacement:
\[ h = \frac{1}{2} g t^2 \]
Where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). Rearranging the formula to solve for \( t \):
\[ t^2 = \frac{2h}{g} \implies t = \sqrt{\frac{2h}{g}} \]
Substituting the values:
\[ t = \sqrt{\frac{2 \times 1960}{9.8}} \approx \sqrt{\frac{3920}{9.8}} \approx \sqrt{400} = 20 \, \text{s} \]

### Step 4: Calculate Horizontal Distance \( AB \)
Now that we have the time of flight, we can calculate the horizontal distance \( AB \) using the formula:
\[ AB = v \times t \]
Substituting the values:
\[ AB = 166.67 \, \text{m/s} \times 20 \, \text{s} = 3333.33 \, \text{m} \]

### Step 5: Convert Distance to Kilometers
To express the distance in kilometers:
\[ AB = \frac{3333.33 \, \text{m}}{1000} \approx 3.33 \, \text{km} \]

### Final Answer
The distance \( AB \) is approximately \( 3.33 \, \text{km} \).

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