The initial speed of a bullet fired from a rifle is 630 m/s. . The rifle is fired at the centre of a target 700 m away at the same level as the target. How far above the centre of target, the rifle must be aimed in order to hit the target?

Options

A. (a)6.1 m
B. (b)1.0 m
C. (c) 9.8 m
D. (d)4.2m

Explanation

To solve the problem of how far above the center of the target the rifle must be aimed in order to hit the target, we can follow these steps:

### Step 1: Understand the problem
We have a bullet fired from a rifle with an initial speed \( u = 630 \, \text{m/s} \) towards a target that is \( 700 \, \text{m} \) away horizontally. We need to determine how high above the center of the target the rifle must be aimed to account for the bullet's drop due to gravity.

### Step 2: Use the range formula for projectile motion
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where:
- \( R \) is the horizontal distance to the target (700 m),
- \( u \) is the initial speed (630 m/s),
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( \theta \) is the angle of projection.

### Step 3: Rearrange the formula to find \( \sin(2\theta) \)
Rearranging the formula gives:
\[ \sin(2\theta) = \frac{R \cdot g}{u^2} \]
Substituting the known values:
\[ \sin(2\theta) = \frac{700 \cdot 10}{630^2} \]

### Step 4: Calculate \( \sin(2\theta) \)
Calculating the right-hand side:
\[ \sin(2\theta) = \frac{7000}{396900} \approx 0.0176 \]

### Step 5: Find \( 2\theta \) using the inverse sine function
Now, we can find \( 2\theta \):
\[ 2\theta = \sin^{-1}(0.0176) \]
Calculating this gives:
\[ 2\theta \approx 1.01^\circ \]
Thus,
\[ \theta \approx 0.505^\circ \]

### Step 6: Calculate the time of flight
To find the time of flight \( t \) until the bullet reaches the target, we can use the horizontal motion:
\[ t = \frac{R}{u \cos(\theta)} \]
Using \( \cos(0.505^\circ) \approx 0.9998 \):
\[ t = \frac{700}{630 \cdot 0.9998} \approx 1.11 \, \text{s} \]

### Step 7: Calculate the vertical drop during time \( t \)
The vertical drop \( h \) due to gravity during time \( t \) is given by:
\[ h = \frac{1}{2} g t^2 \]
Substituting the values:
\[ h = \frac{1}{2} \cdot 10 \cdot (1.11)^2 \approx 6.17 \, \text{m} \]

### Step 8: Conclusion
The rifle must be aimed approximately \( 6.17 \, \text{m} \) above the center of the target to hit it accurately.