A bomb is dropped from an aeroplane flying horizontally with a velocity `469 m s^(-1)` at an altitude of 980 m . The bomb will hit the ground after a time ( use `g = 9.8 m s^(-2)`)

Options

A. 2 s
B. `sqrt2 s`
C. `5sqrt2s`
D. `10sqrt2s`

Answer

To solve the problem of determining the time it takes for a bomb to hit the ground when dropped from an airplane flying horizontally, we can follow these steps:

### Step 1: Identify the known values
- Altitude (height from which the bomb is dropped), \( h = 980 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
- Initial vertical velocity, \( u = 0 \, \text{m/s} \) (since the bomb is dropped)

### Step 2: Use the equation of motion
We will use the second equation of motion for vertical displacement:
\[ S = ut + \frac{1}{2} a t^2 \]
Where:
- \( S \) is the vertical displacement (which will be -980 m, as we consider downward as negative),
- \( u \) is the initial vertical velocity (0 m/s),
- \( a \) is the acceleration (which is -9.8 m/s², as it is acting downward),
- \( t \) is the time in seconds.

Substituting the known values into the equation:
\[ -980 = 0 \cdot t + \frac{1}{2} (-9.8) t^2 \]
This simplifies to:
\[ -980 = -4.9 t^2 \]

### Step 3: Solve for \( t^2 \)
Rearranging the equation gives:
\[ 4.9 t^2 = 980 \]
Now, divide both sides by 4.9:
\[ t^2 = \frac{980}{4.9} \]

### Step 4: Calculate \( t^2 \)
Calculating the right-hand side:
\[ t^2 = 200 \]

### Step 5: Take the square root to find \( t \)
Taking the square root of both sides gives:
\[ t = \sqrt{200} = 10\sqrt{2} \, \text{seconds} \]

### Final Answer
Thus, the bomb will hit the ground after \( t = 10\sqrt{2} \, \text{seconds} \).

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