A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.
At what angle is the projectile fired (measured with respect to the horizontal)?

Options

A. `37^(@)`
B. `45^(@)`
C. `40^(@)`
D. `53^(@)`

Answer

To find the angle at which the projectile is fired with respect to the horizontal, we can use the given horizontal and vertical components of velocity.

### Step-by-Step Solution:

1. **Identify the Components of Velocity**:
- The horizontal component of velocity (\(V_x\)) is given as 30 m/s.
- The vertical component of velocity (\(V_y\)) is given as 40 m/s.

2. **Use the Tangent Function**:
- The angle (\(\theta\)) with respect to the horizontal can be found using the tangent function:
\[ \tan(\theta) = \frac{V_y}{V_x} \]
- Substituting the values:
\[ \tan(\theta) = \frac{40 \, \text{m/s}}{30 \, \text{m/s}} = \frac{4}{3} \]

3. **Calculate the Angle**:
- To find the angle \(\theta\), take the arctangent (inverse tangent) of \(\frac{4}{3}\):
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

4. **Use a Calculator to Find the Angle**:
- Using a calculator, we find:
\[ \theta \approx 53.13^\circ \]

5. **Conclusion**:
- The projectile is fired at an angle of approximately \(53.13^\circ\) with respect to the horizontal.

### Final Answer:
The angle at which the projectile is fired is approximately \(53.13^\circ\).

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