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

Explanation

To solve the problem, we need to determine the height above the target at which the bullet must be aimed in order to hit a target that is 400 m away, given a muzzle velocity of 400 m/s and gravitational acceleration of 10 m/s².

### Step-by-Step Solution:

1. **Understand the Problem**:
- A bullet is fired with an initial velocity (muzzle velocity) of \( u = 400 \, \text{m/s} \).
- The horizontal distance to the target is \( R = 400 \, \text{m} \).
- We need to find the height \( h \) above the target where the bullet should be aimed.

2. **Use the Range Formula**:
- The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
- Here, \( g = 10 \, \text{m/s}^2 \).

3. **Substitute Known Values**:
- Substitute the known values into the range formula:
\[ 400 = \frac{(400)^2 \sin(2\theta)}{10} \]

4. **Simplify the Equation**:
- Rearranging the equation gives:
\[ 400 = \frac{160000 \sin(2\theta)}{10} \]
- Simplifying further:
\[ 400 = 16000 \sin(2\theta) \]
- Dividing both sides by 16000:
\[ \sin(2\theta) = \frac{400}{16000} = \frac{1}{40} \]

5. **Find \( 2\theta \)**:
- We can use the small angle approximation where \( \sin(2\theta) \approx 2\theta \) for small angles:
\[ 2\theta \approx \frac{1}{40} \]
- Therefore,
\[ \theta \approx \frac{1}{80} \, \text{radians} \]

6. **Calculate the Height \( h \)**:
- For small angles, \( \tan(\theta) \approx \theta \):
\[ \tan(\theta) = \frac{h}{R} \]
- Substituting \( R = 400 \, \text{m} \):
\[ \frac{1}{80} = \frac{h}{400} \]

7. **Solve for \( h \)**:
- Rearranging gives:
\[ h = 400 \times \frac{1}{80} = 5 \, \text{m} \]

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

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