An aeroplane is flying in a horizontal direction with a velocity `600 km//h` at a height of 1960 m. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance AB.

Options

A. `3.33 km`
B. `4.33km`
C. `5.33km`
D. `6.33km`

Explanation

To solve the problem of finding the distance AB where a body dropped from an aeroplane strikes the ground, we will follow these steps:

### Step 1: Convert the velocity from km/h to m/s
The velocity of the airplane is given as 600 km/h. We need to convert this to meters per second (m/s) using the conversion factor \(1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s}\).

\[ \text{Velocity in m/s} = 600 \, \text{km/h} \times \frac{1 \, \text{m/s}}{3.6 \, \text{km/h}} = \frac{600}{3.6} \approx 166.67 \, \text{m/s} \]

### Step 2: Calculate the time of flight
The body is dropped from a height of 1960 m. We can use the equation of motion to calculate the time it takes for the body to fall to the ground. The equation is:

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

Where:
- \(s = 1960 \, \text{m}\) (height)
- \(u = 0 \, \text{m/s}\) (initial vertical velocity)
- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)

Substituting the values:

\[ 1960 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]

This simplifies to:

\[ 1960 = 4.9t^2 \]

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

\[ t^2 = \frac{1960}{4.9} \approx 400 \]

Taking the square root:

\[ t = \sqrt{400} = 20 \, \text{s} \]

### Step 3: Calculate the horizontal distance (range)
Now that we have the time of flight, we can calculate the horizontal distance (distance AB) using the formula:

\[ \text{Distance} = \text{Velocity} \times \text{Time} \]

Substituting the values:

\[ \text{Distance} = 166.67 \, \text{m/s} \times 20 \, \text{s} = 3333.4 \, \text{m} \]

Converting this to kilometers:

\[ \text{Distance} = \frac{3333.4}{1000} \approx 3.33 \, \text{km} \]

### Final Answer
The distance AB is approximately **3.33 km**.

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