An aeroplane moving horizontally with a speed of 180km/hr. drops a food packet while flying at a height of 490m. The horizontal range of the packet is:-

Options

A. 180m
B. 980m
C. 500m
D. 670m

Explanation

To solve the problem, we need to calculate the horizontal range of the food packet dropped from the aeroplane. Here are the steps to find the solution:

### Step 1: Convert the speed of the aeroplane from km/hr to m/s.
The speed of the aeroplane is given as 180 km/hr. To convert this to m/s, we use the conversion factor:
\[ 1 \text{ km/hr} = \frac{1}{3.6} \text{ m/s} \]
Thus,
\[ \text{Speed in m/s} = 180 \times \frac{1}{3.6} = 50 \text{ m/s} \]

### Step 2: Calculate the time taken for the packet to fall to the ground.
The height from which the packet is dropped is 490 m. We can use the formula for the time of free fall:
\[ h = \frac{1}{2} g t^2 \]
Where:
- \( h = 490 \) m (height)
- \( g = 9.8 \) m/s² (acceleration due to gravity)

Rearranging the formula to solve for \( t \):
\[ t^2 = \frac{2h}{g} = \frac{2 \times 490}{9.8} \]
Calculating:
\[ t^2 = \frac{980}{9.8} = 100 \quad \Rightarrow \quad t = \sqrt{100} = 10 \text{ seconds} \]

### Step 3: Calculate the horizontal range of the packet.
The horizontal range \( R \) can be calculated using the formula:
\[ R = u \times t \]
Where:
- \( u = 50 \) m/s (horizontal speed of the aeroplane)
- \( t = 10 \) seconds (time of fall)

Substituting the values:
\[ R = 50 \times 10 = 500 \text{ m} \]

### Conclusion:
The horizontal range of the packet is **500 meters**.

---