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 how long it takes for a bomb dropped from an airplane to hit the ground, we can use the equations of motion under gravity. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Initial height (h) = 980 m
- Acceleration due to gravity (g) = 9.8 m/s²
- Initial vertical velocity (u_y) = 0 m/s (since the bomb is dropped)

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

### Step 3: Plug in the values
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 4: Rearrange the equation
To isolate \( t^2 \), we can rearrange the equation:
\[ t^2 = \frac{980}{4.9} \]

### Step 5: Calculate \( t^2 \)
Now, calculate \( t^2 \):
\[ t^2 = 200 \]

### Step 6: Take the square root to find \( t \)
Now, take the square root to find \( t \):
\[ t = \sqrt{200} = 10\sqrt{2} \approx 14.14 \text{ seconds} \]

### Conclusion
The bomb will hit the ground after approximately 14.14 seconds.

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