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 solve the problem of finding the projection angle with the horizontal for a projectile, we will follow these steps:

### Step 1: Understand the given information
We are given:
- Time of flight (T) = 1 second
- Range (R) = 4 meters

### Step 2: Use the time of flight formula
The time of flight for a projectile is given by the formula:
\[ T = \frac{2u \sin \theta}{g} \]
Where:
- \( u \) is the initial velocity
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))

Rearranging the formula to find \( u \sin \theta \):
\[ u \sin \theta = \frac{gT}{2} \]
Substituting the values:
\[ u \sin \theta = \frac{10 \times 1}{2} = 5 \, \text{m/s} \]

### Step 3: Use the range formula
The range of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Rearranging this formula gives us:
\[ u^2 \sin 2\theta = Rg \]
Substituting the known values:
\[ u^2 \sin 2\theta = 4 \times 10 = 40 \]

### Step 4: Express \( \sin 2\theta \) in terms of \( u \)
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we can express this as:
\[ u^2 (2 \sin \theta \cos \theta) = 40 \]
Substituting \( u \sin \theta = 5 \):
\[ u^2 \cdot 2 \cdot \frac{5}{u} \cdot \cos \theta = 40 \]
This simplifies to:
\[ 10u \cos \theta = 40 \]
Thus,
\[ u \cos \theta = 4 \]

### Step 5: Solve for \( u \)
Now we have two equations:
1. \( u \sin \theta = 5 \)
2. \( u \cos \theta = 4 \)

### Step 6: Find \( u \)
To find \( u \), we can use the Pythagorean theorem:
\[ u = \sqrt{(u \sin \theta)^2 + (u \cos \theta)^2} \]
Substituting the values:
\[ u = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \]

### Step 7: Find \( \sin \theta \) and \( \cos \theta \)
Now we can find \( \sin \theta \) and \( \cos \theta \):
- From \( u \sin \theta = 5 \):
\[ \sin \theta = \frac{5}{u} = \frac{5}{\sqrt{41}} \]
- From \( u \cos \theta = 4 \):
\[ \cos \theta = \frac{4}{u} = \frac{4}{\sqrt{41}} \]

### Step 8: Find \( \tan \theta \)
Now we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{5}{\sqrt{41}}}{\frac{4}{\sqrt{41}}} = \frac{5}{4} \]

### Step 9: Calculate \( \theta \)
Finally, we can find the angle \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{5}{4}\right) \]

### Final Answer
The projection angle with the horizontal is \( \theta = \tan^{-1}\left(\frac{5}{4}\right) \).

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