A rifle shoots a bullet with a muzzle velocity of `400 m s^-1` at a small target `400 m` away. The height above the target at which the bullet must be aimed to hit the target is `(g = 10 m s^-2)`.

Options

A. 1 m
B. 5 m
C. 10 m
D. 0.5 m

Explanation

To solve the problem of how high above the target the bullet must be aimed to hit a target 400 m away, we can follow these steps:

### Step 1: Understand the Problem
We need to find the height \( h \) above the target where the rifle must be aimed. The bullet is fired with a muzzle velocity \( u = 400 \, \text{m/s} \) and the target is at a horizontal distance \( R = 400 \, \text{m} \).

### Step 2: Use the Range Formula
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity. In this case, \( g = 10 \, \text{m/s}^2 \).

### Step 3: Substitute Known Values
Substituting the known values into the range formula:
\[ 400 = \frac{(400)^2 \sin(2\theta)}{10} \]
This simplifies to:
\[ 400 = \frac{160000 \sin(2\theta)}{10} \]
\[ 400 = 16000 \sin(2\theta) \]
Dividing both sides by 16000 gives:
\[ \sin(2\theta) = \frac{400}{16000} = \frac{1}{40} \]

### Step 4: Approximate for Small Angles
For small angles, we can use the approximation \( \sin(2\theta) \approx 2\theta \). Therefore:
\[ 2\theta \approx \frac{1}{40} \]
This implies:
\[ \theta \approx \frac{1}{80} \, \text{radians} \]

### Step 5: Relate Height to Angle
The height \( h \) above the target can be related to the angle \( \theta \) using the tangent function:
\[ \tan(\theta) = \frac{h}{R} \]
For small angles, \( \tan(\theta) \approx \theta \). Thus:
\[ \frac{h}{400} \approx \frac{1}{80} \]

### Step 6: Solve for Height \( h \)
Now, we can solve for \( h \):
\[ h \approx 400 \cdot \frac{1}{80} = 5 \, \text{m} \]

### Conclusion
The height above the target at which the bullet must be aimed is \( h = 5 \, \text{m} \).