An arrow is shot in air, its time of flight is 5 sec and horizontal range is 200 m. the inclination of the arrow with the horizontal is

Options

A. `tan^(-1).(5)/(8)`
B. `tan^(-1).(8)/(5)`
C. `tan^(-1).(1)/(8)`
D. `45^(@)`

Explanation

To solve the problem of finding the angle of inclination of the arrow with the horizontal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Information**:
- Time of flight (T) = 5 seconds
- Horizontal range (R) = 200 meters

2. **Understand the Components of Motion**:
- When the arrow is shot at an angle θ, its initial velocity (V) can be resolved into two components:
- Horizontal component: \( V_x = V \cos \theta \)
- Vertical component: \( V_y = V \sin \theta \)

3. **Use the Horizontal Range Formula**:
- The horizontal range (R) can be expressed as:
\[ R = V_x \cdot T \]
- Substituting the horizontal component:
\[ R = (V \cos \theta) \cdot T \]
- Plugging in the values:
\[ 200 = (V \cos \theta) \cdot 5 \]
- Rearranging gives:
\[ V \cos \theta = \frac{200}{5} = 40 \quad \text{(Equation 1)} \]

4. **Use the Time of Flight Formula**:
- The time of flight (T) for projectile motion is given by:
\[ T = \frac{2 V_y}{g} \]
- Substituting the vertical component:
\[ T = \frac{2 (V \sin \theta)}{g} \]
- Plugging in the values (where g = 10 m/s²):
\[ 5 = \frac{2 (V \sin \theta)}{10} \]
- Rearranging gives:
\[ V \sin \theta = 25 \quad \text{(Equation 2)} \]

5. **Divide Equation 2 by Equation 1**:
- To eliminate V, we divide Equation 2 by Equation 1:
\[ \frac{V \sin \theta}{V \cos \theta} = \frac{25}{40} \]
- This simplifies to:
\[ \tan \theta = \frac{25}{40} = \frac{5}{8} \]

6. **Calculate the Angle θ**:
- To find θ, take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{5}{8}\right) \]

7. **Final Answer**:
- The angle of inclination of the arrow with the horizontal is:
\[ \theta = \tan^{-1}\left(\frac{5}{8}\right) \]