Projection angle with the horizontal is:

Options

A. `tan^(-1)(4/5)`
B. `tan^(-1)(2/3)`
C. `tan^(-1)(5/4)`
D. `tan^(-1)(1/2)`

Explanation

To find the projection angle with the horizontal given the time of flight and range, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Time of flight (T) = 1 second
- Range (R) = 4 meters
- Acceleration due to gravity (g) = 10 m/s² (approximately)

2. **Use the formula for time of flight:**
The time of flight for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
Substituting the values we have:
\[ 1 = \frac{2u \sin \theta}{10} \]
Rearranging gives:
\[ u \sin \theta = \frac{10}{2} = 5 \text{ m/s} \]
(Equation 1)

3. **Use the formula for range:**
The range of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
We can also express \(\sin 2\theta\) as \(2 \sin \theta \cos \theta\):
\[ R = \frac{u^2 \cdot 2 \sin \theta \cos \theta}{g} \]
Substituting the values:
\[ 4 = \frac{u^2 \cdot 2 \sin \theta \cos \theta}{10} \]
Rearranging gives:
\[ u^2 \sin \theta \cos \theta = \frac{4 \cdot 10}{2} = 20 \text{ m} \]
(Equation 2)

4. **Relate \(u \sin \theta\) and \(u \cos \theta\):**
From Equation 1, we have \(u \sin \theta = 5\). Now we need to find \(u \cos \theta\):
\[ R = u \cos \theta \cdot T \]
Substituting the known values:
\[ 4 = u \cos \theta \cdot 1 \]
Thus:
\[ u \cos \theta = 4 \text{ m/s} \]
(Equation 3)

5. **Divide Equation 1 by Equation 3 to find \(\tan \theta\):**
\[ \frac{u \sin \theta}{u \cos \theta} = \tan \theta \]
Substituting the values from Equations 1 and 3:
\[ \tan \theta = \frac{5}{4} \]

6. **Calculate the angle \(\theta\):**
To find the angle \(\theta\), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{5}{4}\right) \]

7. **Conclusion:**
The projection angle with the horizontal is:
\[ \theta = \tan^{-1}\left(\frac{5}{4}\right) \]