A body is projected at an angle of `30^@` with the horizontal and with a speed of `30 ms^-1`. What is the angle with the horizontal after `1.5 s` ? `(g = 10 ms^-2)`.

Options

A. `0^@`
B. `30^@`
C. `60^@`
D. `90^@`

Explanation

To solve the problem, we need to find the angle with the horizontal of a body projected at an angle of \(30^\circ\) with a speed of \(30 \, \text{m/s}\) after \(1.5 \, \text{s}\).

### Step-by-Step Solution:

1. **Identify the Initial Velocity Components**:
The initial velocity \(u\) is given as \(30 \, \text{m/s}\) and the angle of projection \(\theta\) is \(30^\circ\).
- The horizontal component of the velocity (\(u_x\)) is given by:
\[ u_x = u \cos \theta = 30 \cos 30^\circ = 30 \times \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \text{m/s} \]
- The vertical component of the velocity (\(u_y\)) is given by:
\[ u_y = u \sin \theta = 30 \sin 30^\circ = 30 \times \frac{1}{2} = 15 \, \text{m/s} \]

2. **Calculate the Vertical Velocity After 1.5 Seconds**:
The vertical velocity (\(v_y\)) at time \(t\) can be calculated using the formula:
\[ v_y = u_y - gt \]
where \(g\) is the acceleration due to gravity (\(10 \, \text{m/s}^2\)).
- Substituting the values:
\[ v_y = 15 - 10 \times 1.5 = 15 - 15 = 0 \, \text{m/s} \]

3. **Determine the Angle with the Horizontal**:
The angle \(\phi\) with the horizontal can be found using the tangent of the angle:
\[ \tan \phi = \frac{v_y}{u_x} \]
Since \(v_y = 0\):
\[ \tan \phi = \frac{0}{15\sqrt{3}} = 0 \]
This implies that:
\[ \phi = \tan^{-1}(0) = 0^\circ \]

### Conclusion:
After \(1.5 \, \text{s}\), the angle with the horizontal is \(0^\circ\).