A bomb is dropped from an aeroplane when it is at a height h directly above a target. If the aeroplane is moving horizontally at a speed v, the distance by which the bomb will miss the target is given by

Options

A. `2V sqrt(h/g)`
B. `v sqrt(h/g)`
C. `vsqrt((2h)/g)`
D. `vsqrt(h/(2g))`

Explanation

To solve the problem of how far the bomb will miss the target when dropped from an airplane moving horizontally, we can break it down into a few steps:

### Step 1: Understand the scenario
The bomb is dropped from a height \( h \) while the airplane is moving horizontally with a speed \( v \). The bomb will fall under the influence of gravity and will also have the horizontal speed of the airplane.

### Step 2: Calculate the time of flight
To find out how long the bomb is in the air, we can use the second equation of motion for vertical motion. The equation is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
For the vertical motion:
- \( s = h \) (the height from which the bomb is dropped),
- \( u = 0 \) (initial vertical velocity),
- \( a = g \) (acceleration due to gravity).

Substituting these values into the equation:
\[ h = 0 \cdot t + \frac{1}{2} g t^2 \]
This simplifies to:
\[ h = \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t^2 = \frac{2h}{g} \]
Taking the square root:
\[ t = \sqrt{\frac{2h}{g}} \]

### Step 3: Calculate the horizontal distance traveled
The horizontal distance \( S \) that the bomb travels while falling can be calculated using the formula:
\[ S = vt \]
Substituting the expression for \( t \):
\[ S = v \cdot \sqrt{\frac{2h}{g}} \]

### Conclusion
The distance by which the bomb will miss the target is given by:
\[ S = v \sqrt{\frac{2h}{g}} \]