An airplane moving horizontally with a speed of `18km//hr` drops a food packet while flying at a height of 500m. The horizontal range is:

Explanation

To find the horizontal range of the food packet dropped from the airplane, we can follow these steps:

### Step 1: Convert the speed of the airplane from km/hr to m/s
The speed of the airplane is given as 18 km/hr. To convert this to meters per second (m/s), we use the conversion factor:
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
Thus,
\[ \text{Speed in m/s} = 18 \times \frac{5}{18} = 5 \text{ m/s} \]

### Step 2: Identify the height from which the packet is dropped
The height from which the food packet is dropped is given as 500 m.

### Step 3: Calculate the time of flight
To find the time it takes for the packet to fall 500 m, we can use the formula for the distance fallen under gravity:
\[ h = \frac{1}{2} g t^2 \]
Where:
- \( h = 500 \) m (height)
- \( g = 10 \) m/s² (acceleration due to gravity)

Rearranging the formula to solve for \( t \):
\[ t^2 = \frac{2h}{g} \]
Substituting the known values:
\[ t^2 = \frac{2 \times 500}{10} = 100 \]
Taking the square root:
\[ t = \sqrt{100} = 10 \text{ seconds} \]

### Step 4: Calculate the horizontal range
The horizontal range \( R \) can be calculated using the formula:
\[ R = u \cdot t \]
Where:
- \( u = 5 \) m/s (horizontal speed of the packet)
- \( t = 10 \) seconds (time of flight)

Substituting the values:
\[ R = 5 \cdot 10 = 50 \text{ meters} \]

### Final Answer
The horizontal range of the food packet is **50 meters**.

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