An aeroplane is flying in a horizontal direction with a velocityu and at a height of 2000 m. When it is vertically below a point A on the ground a food packet is released from it. The packet strikes the ground at point B. If AB = 3 km and g =10 `m//s^(2)` , then the value of u is

Options

A. 54 km/h
B. 540 km/h
C. 150 km/h
D. 300 km/h

Explanation

To solve the problem, we need to find the horizontal velocity \( u \) of the aeroplane when a food packet is released from a height of 2000 m and strikes the ground at a distance of 3 km horizontally. We will use the principles of projectile motion to derive the solution step-by-step.

### Step 1: Identify the given values
- Height \( h = 2000 \, \text{m} \)
- Horizontal distance \( AB = 3 \, \text{km} = 3000 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the time of flight \( t \)
The time taken for the packet to fall from the height \( h \) can be calculated using the formula for free fall:
\[ h = \frac{1}{2} g t^2 \]
Rearranging the formula to solve for \( t \):
\[ t^2 = \frac{2h}{g} \]
Substituting the values:
\[ t^2 = \frac{2 \times 2000}{10} = \frac{4000}{10} = 400 \]
Taking the square root:
\[ t = \sqrt{400} = 20 \, \text{s} \]

### Step 3: Use the range formula to find \( u \)
The horizontal distance (range) covered by the packet can be expressed as:
\[ \text{Range} = u \cdot t \]
Substituting the known values:
\[ 3000 = u \cdot 20 \]
Solving for \( u \):
\[ u = \frac{3000}{20} = 150 \, \text{m/s} \]

### Step 4: Convert \( u \) to km/h (if required)
To convert from meters per second to kilometers per hour, we use the conversion factor \( \frac{18}{5} \):
\[ u = 150 \, \text{m/s} \times \frac{18}{5} = 540 \, \text{km/h} \]

### Final Answer
The horizontal velocity \( u \) of the aeroplane is \( 150 \, \text{m/s} \) or \( 540 \, \text{km/h} \).

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