A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of `60 km//hr` and at a height of 490 m. How far the aeroplane must be from the enemy post at time of dropping the bomb, so that it may directly hit the target ?` (g = 9.8 m//s^(2)).` What is the trajectory of the bomb as seen by an observer on the earth? What as seen by a person sitting inside the aeroplane?

Options

A. `(400)/3m`
B. `(500)/3 m`
C. `(1700)/3m`
D. `498 m`

Explanation

To solve the problem of how far the aeroplane must be from the enemy post at the time of dropping the bomb, we can break it down into a series of steps:

### Step 1: Convert the horizontal velocity from km/hr to m/s
The horizontal velocity of the aeroplane is given as 60 km/hr. We need to convert this to meters per second (m/s) using the conversion factor:

\[ 1 \text{ km/hr} = \frac{1}{3.6} \text{ m/s} \]

So,

\[ 60 \text{ km/hr} = 60 \times \frac{1}{3.6} \text{ m/s} = \frac{60}{3.6} \text{ m/s} = \frac{600}{36} \text{ m/s} = \frac{50}{3} \text{ m/s} \approx 16.67 \text{ m/s} \]

### Step 2: Calculate the time taken for the bomb to hit the ground
The height from which the bomb is dropped is 490 m. We can use the second equation of motion to find the time taken (t) to hit the ground. The equation is:

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

Where:
- \(s\) = height (490 m)
- \(u\) = initial vertical velocity (0 m/s, since the bomb is dropped)
- \(g\) = acceleration due to gravity (9.8 m/s²)

Substituting the values, we have:

\[ 490 = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 \]

This simplifies to:

\[ 490 = 4.9 t^2 \]

Solving for \(t^2\):

\[ t^2 = \frac{490}{4.9} = 100 \]

Taking the square root:

\[ t = 10 \text{ seconds} \]

### Step 3: Calculate the horizontal distance traveled by the bomb
Now that we have the time taken for the bomb to hit the ground, we can calculate the horizontal distance (d) it travels during this time. Since there is no horizontal acceleration, we can use the formula:

\[ d = u \cdot t \]

Where:
- \(u\) = horizontal velocity of the bomb (which is the same as the aeroplane's velocity, \(\frac{50}{3} \text{ m/s}\))
- \(t\) = time (10 seconds)

Substituting the values:

\[ d = \frac{50}{3} \cdot 10 = \frac{500}{3} \text{ meters} \approx 166.67 \text{ meters} \]

### Step 4: Determine the trajectory of the bomb
- **As seen by an observer on the Earth:** The trajectory of the bomb will appear to be a parabolic path, as it is a projectile motion with an initial horizontal velocity and is acted upon by gravity.

- **As seen by a person sitting inside the aeroplane:** The bomb will appear to fall straight down vertically. This is because the bomb has the same horizontal velocity as the aeroplane, making it seem stationary in the horizontal direction from the perspective of someone inside the plane.

### Final Answer:
The aeroplane must be approximately \( \frac{500}{3} \) meters (or about 166.67 meters) away from the enemy post at the time of dropping the bomb.

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