An aeroplane flying horizontally at an altitude of 490m with a speed of 180 kmph drops a bomb. The horizontal distance at which it hits the ground is

Options

A. 500 m
B. 1000 m
C. 250 m
D. 50 m

Answer

To solve the problem of finding the horizontal distance at which a bomb dropped from an airplane hits the ground, we will follow these steps:

### Step 1: Identify the given data
- Altitude (h) = 490 m
- Speed of the airplane (u) = 180 km/h

### Step 2: Convert speed from km/h to m/s
To convert the speed from kilometers per hour to meters per second, we use the conversion factor:
\[ 1 \text{ km/h} = \frac{5}{18} \text{ m/s} \]
Thus,
\[ u = 180 \text{ km/h} \times \frac{5}{18} = 50 \text{ m/s} \]

### Step 3: Calculate the time of flight (T)
The time taken for the bomb to hit the ground can be calculated using the formula for free fall:
\[ T = \sqrt{\frac{2h}{g}} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)). Substituting the values:
\[ T = \sqrt{\frac{2 \times 490}{9.8}} = \sqrt{\frac{980}{9.8}} = \sqrt{100} = 10 \text{ seconds} \]

### Step 4: Calculate the horizontal distance (R)
The horizontal distance (range) can be calculated using the formula:
\[ R = u \times T \]
Substituting the values we found:
\[ R = 50 \text{ m/s} \times 10 \text{ s} = 500 \text{ m} \]

### Conclusion
The horizontal distance at which the bomb hits the ground is **500 meters**.

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