An aeroplane moving horizontally with a speed of `720 km//h` drops a food pocket, while flying at a height of 396.9 m . the time taken by a food pocket to reach the ground and its horizontal range is (Take `g = 9.8 m//sec`)

Options

A. `3 sec` and `2000m`
B. `5 sec` and `500m`
C. `8 sec` and `1500 m`
D. `9 sec` and `1800m`

Answer

To solve the problem step by step, we need to find two things: the time taken by the food packet to reach the ground and its horizontal range.

### Step 1: Convert the speed of the airplane from km/h to m/s
The speed of the airplane is given as \(720 \text{ km/h}\). We need to convert this to meters per second (m/s).

\[ \text{Speed in m/s} = \frac{720 \times 1000 \text{ m}}{3600 \text{ s}} = \frac{720000 \text{ m}}{3600 \text{ s}} = 200 \text{ m/s} \]

### Step 2: Calculate the time taken to reach the ground
We will use the second equation of motion to find the time taken for the food packet to fall from a height of \(396.9 \text{ m}\):

\[ s = ut + \frac{1}{2}gt^2 \]

Where:
- \(s = 396.9 \text{ m}\) (the height)
- \(u = 0 \text{ m/s}\) (initial vertical velocity)
- \(g = 9.8 \text{ m/s}^2\) (acceleration due to gravity)

Substituting the values into the equation:

\[ 396.9 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]

This simplifies to:

\[ 396.9 = 4.9t^2 \]

Now, solving for \(t^2\):

\[ t^2 = \frac{396.9}{4.9} \approx 81 \]

Taking the square root to find \(t\):

\[ t = \sqrt{81} = 9 \text{ seconds} \]

### Step 3: Calculate the horizontal range
The horizontal range can be calculated using the formula:

\[ \text{Range} = \text{horizontal speed} \times \text{time} \]

The horizontal speed is \(200 \text{ m/s}\) and the time taken is \(9 \text{ seconds}\):

\[ \text{Range} = 200 \text{ m/s} \times 9 \text{ s} = 1800 \text{ m} \]

### Final Answers
- Time taken to reach the ground: \(9 \text{ seconds}\)
- Horizontal range: \(1800 \text{ meters}\)

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